1

Ch2

2.1

Notice: in general, average speed for the total trip is not the average of speeds for each leg of the trip.

Notice: if an object changes direction during motion, distance is not the absolute value of displacement.

Notice: if an object changes direction during motion, average speed is not the absolute value of average velocity.

Comment 1: Typically we keep 3 sig figs on all numbers unless otherwise specified. Here I got lazy since I wanted

to focus on when you should use � ̂and when you shouldn’t.

Comment 2: Typically, any time you write a number (instead of a variable) you should include appropriate units.

In this case I did include units for each number in the column headings. This is a useful trick for data tables.

time interval ∆� ( m ) ∆�� ( m ) � (�� ) �� (�� )

0-4 sec 10 10 �̂ 2.5 2.5 �̂ 4-5 sec 15 -15 �̂ 15 -15 �̂ 5-6 sec 5 5 �̂ 5 5 �̂

total trip

(0-6 sec) 30 0 �̂ ��������������� = 30m6s = 5ms

����� !���"�������� = 0m�̂6s = 0

2

2.1¼

a) NO; # is the magnitude of freefall acceleration near earth’s surface.

b) NO; # is the magnitude of freefall acceleration near earth’s surface.

c) NO; # is the magnitude of freefall acceleration near earth’s surface.

This magnitude must always be positive.

We will always use # = $9.8 ��(. d) First make the picture shown at right.

Next, make the list of knowns shown in the table below the picture.

Next list out the big three equations.

�) = �* $ �*+ $ 12 �+. �)+ = �*+ $ �+ �)+. = �*+. $ 2�+Δ�

Next, THINK! Which of the above equations helps us determine

our desired outcome? Problem asked for distance traveled. Third

equation will get us there.

Now, do the problem with algebra before plugging in numbers.

Tip: to make the algebra simpler…DO plug in any zeroes from the

table at right. 0 = �*+. $ 2�+Δ� 0�*+. = 2�+Δ�

WATCH OUT! You are supposed to know 0�*+. = 01�*+. 2 from

order of operations!!! To be painfully clear, in this instance the

squaring of �*+ does not get rid of the minus sign!!!

Δ� = 0 �*+.2�+

Check the units! Tip: a minus sign doesn’t affect the units…

3Δ�4 = 3�*+. 43243�+4 =5ms 6.ms. = m.s. ∙ s.m = m

Units look good. Now plug in numbers.

Tip: since we already checked the units, I could leave the units off when plugging in numbers!!!

Tip: include units on your final answer…otherwise other people won’t know if you mean miles or meters.

Δ� = 0 575.98ms 6.2 501.715 ms.6 = 1683m = 1.046mi Notice the minus sign on �+ makes sense! Also, seems reasonable for a landing distance.

Tip: if the first digit is a 1, some engineers include an extra sig fig (beyond the standard 3) in the result.

WATCH OUT! We solved for displacement, not distance. These kinematic equations assume � ̂is implied.

Fortunately, the plane never reversed direction. In this case, magnitude of displacement is distance.

e) Let’s get the time using �)+ = �*+ $ �+ 0 = �*+ $ �+ = 0 �*+�+ = 0 575.98ms 6501.715 ms.6 = 44.3s

Checks: units look good, time seems reasonable.

�* 0 (assume origin at start) �) ?

�*+

170mihr > 1609m1mi > 1hr3600s = 75.98ms

Note: strictly speaking we should assume 2

sig figs on 170. That said, as engineers you

are expected to compute assuming 3 sig figs

for final answer unless otherwise specified.

That means we should keep 4 on this number

to avoid intermediate rounding error. �)+ 0 (comes to rest)

�+

00.175# = 00.17559.8ms.6 = 01.715ms.

The minus here comes from two things:

1) We chose “to the right” as $� direction.

2) The acceleration points opposite $�. ?

$�

$? �@ �) = 0

��#A�BC� = 0.175#

3

2.1½

Parts a & b solved together: My figure looks about the same as last time.

Table of knowns and unknowns is again shown.

Big three equations are again:

�) = �* $ �*+ $ 12 �+. �)+ = �*+ $ �+ �)+. = �*+. $ 2�+Δ�

Next, THINK! Which of the above equations helps us determine our desired

outcome? Problem asked for time. Two methods come to mind:

1) Use the quadratic formula with the first equation. A great method.

2) First use the third equation to solve for �+ (asked for in part b) then

use the 2nd equation to solve for . I will use the second method so I can use more cut and paste… 0 = �*+. $ 2�+Δ� 0�*+. = 2�+Δ�

�+ = 0 �*+.2Δ�

�+ = 0535.0ms 6.

219.00m2 = 068. 06 ms.

Units check.

Direction makes sense (negative sign on � because moving to the right but slowing down).

This acceleration seems large but I will discuss that in a minute…

I keep an extra sig fig. Why? I plan to use this number to get time and I want to avoid intermediate rounding error. �)+ = �*+ $ �+ 0 = �*+ $ �+ = 0 �*+�+ = 0 535.0ms 65068. 06 ms.6 = 0.514s

Checks: units look good, time seems reasonable.

Part c) Since we did the problem algebraically, we simply change the numbers in the last lines of our work!

If the distance is now 1.50m instead of 9.00m one finds �+ = 0408. 3 ms.

Part d) Using a 9.00m stopping distance we got �+ = 06.94# D 07#.

Using a 1.50m stopping distance we got �+ = 41.7# D 40#.

Notice: by solving the problem algebraically, it is easy to check other scenarios after the fact…

An acceleration magnitude of 7# is probably survivable for must humans while the 40# is not.

Tip: if you drive the speed limit (let’s assume 65mph D 30�� ) the acceleration magnitude for a 9.00m stopping

distance drops to only 5# and you are significantly more likely to live.

�* 0 (assume origin at start)

�) 9.00m

�*+ 35.0ms

�)+ 0 (comes to rest)

�+ ? ?

$�

$? �@ �) = 0

�+ =?

4

2.1¾ The work for both parts is shown.

My figure is now getting more complicated.

This problem has two stages.

I make two separate lists of knowns!!!

Now we see the importance of labelling the figure clearly.

Side note: You might be able to do all the work in your head, but it

is much easier to label things clearly. Just think, what if a problem had

three or four stages…your head gets filled up too quickly. Better to do

things on paper and not have to think so hard while doing

computations.

We continue with the same procedure as before, only this time we set

up two sets of equations. See the table below.

Work for Stage 1

�F) = �F* $ �F*+F $ 12�F+F. �F)+ = �F*+ $ �F+F �F)+. = �F*+. $ 2�F+Δ�F

First plug in all zeroes and rewrite.

�F) = 12�F+F. �F)+ = �F+F �F)+. = 2�F+Δ�F

Now think: the last equation can get me �GHI. The first equation can get me time. If desired, I could then use the 2nd

equation to check my math! I found �F)+ = �GHI = J2�F+Δ�F = 11.94ms

Tip: in general, always consider both K roots when using the square root function. Sometimes the negative root is the

one that makes sense!!! In this instance, the runner is going in the positive direction…use $.

I checked the units. I converted this speed to miles per hour to do a reality check (11.94�� D 27mph‼!) pretty darn

fast but we were told the woman is a “track star”. Seems ok.

I found the time for this stage as well.

F = N2�F)�F+ = 2.513s Units check. Time seems plausible.

Work for Stage 2

Tip: final position from stage 1 is initial position for stage 2. Final velocity from stage 1 is initial velocity stage 2.

To save some space, I will plug in the zero’s into the big three equations right away. The first equation becomes �.) = �.* $ �.*+.

. = �.) 0 �.*�.*+ = 7.119s

TOTAL TIME is thus ��O��! = �P $ �Q = R. STU…A NEW WORLD RECORD!!!

�F* 0 �.* 15.0m �F) 15.0m �.) 100m

�F*+ 0 (starts from

rest) �.*+ �GHI

�F)+ �GHI �.)+ �GHI

�F+ 4.75 ms. �.+ 0 F ? . ?

$� �F*+ = 0

4.75ms.

�F)+ = �.*+ = �GHI � = 0

Runner @ constant speed

Total distance is100m

5

2.2

a) List of knowns and unknowns is shown at right. If we assume the angle of the

incline is constant during the block’s motion, we have constant acceleration.

Since acceleration is constant

�) = �* $ �*+ $ 12 �+. �)+ = �*+ + �+ �)+. = �*+. + 2�+1∆�2 Comparing equations to list of knowns, one recognizes the first equation has the

unknown t and the remaining variables are all knowns.

First we plug in the zeros from the list of knowns to clean up the equation. This gives

�) = 12�+.

Then we plug in our knowns giving

V = 12# sin X .

Upon re-reading the question we realize we are looking for X. Solving for X gives

X = sinYF Z 2V#.[

Before moving on we check the units of the right hand side

\2V# ] = 3243V43#43.4 = 3V43#434. = mms. ∙ s. = nounits, asexpectedinsideofsinYF

Reality check: We can also see if the answer makes sense by considering what would happen in some

special cases.

• What happens to X if V is increased? The time increases. If length doubles, the time increases by

a factor of √2. While the factor might surprise you, we do expect increasing V to increase . • What if the ramp was on the moon where acceleration is less? If the acceleration was less, #

would be a smaller number. With a smaller number in the denominator, is longer. This seems to

makes sense: having less acceleration increases time for the block to slide down.

• What if the ramp was parallel to the ground? The angle would be 0°, sin 0° = 0, and = ∞.

This actually seems to make sense as the block remains stationary if the ramp is horizontal.

b) This question now asks us to find a specific angle.

X = sinYF Z 2V#.[

X = sinYF k 212.00m259.8 ms.6 11.40s2.l = 12. 02° Tip: if the first digit of a number is a 1, engineers often will write 4 sig figs instead of 3.

c) 56.4°. To see why we don’t simply double or quadruple the angle, one can rearrange the result from part a

to solve instead for time: = m .no �pqr. We see time does not depend linearly on angle. Doubling the angle

will not, in general cut the time in half. Note: for small angles (θ < 10°) we could use the approximation sin X ≈ X in radians. For X < 5°, doubling the angle doubles the denominator inside the square root.

Therefore, for X < 5°, doubling the angle cuts divides the time by √2. Said another way, for X < 2.5°, quadrupling the angle will effectively cut the travel time in half. If you are interested, sin X ≈ X comes

from the first term of the Taylor series.

Problem continues next page…

xi 0 (assume origin at start)

xf L

vix 0 (“released from rest”)

vfx ?

ax # sin X (shown in figure)

t ?

6

d) When you see a question asking about a max or min it is common to think of taking a derivative. In some

cases, we can consider limiting cases like X = 0° or X = 90° and the answer pops out at you. For instance,

in this case we expect the time should get shorter as one raises the ramp. At 90° we should get a minimum

time.

If you tried to use derivatives you would want to do as follows:

• We want to minimize t with respect to θ.

• Use tGtr = 0 and solve for θcritical.

• That gives CCX = CCX Z 2V# sin X[F/. = Z2V# [

F/. CCX 1sin X2YF/. = Z2V# [F. 1− 1221sin X2Yv/. cos X

• By setting this equal to zero we find

Z2V# [F. Z−12[ 1sin X2Yv. cos X = 0

Z2V# [F. 5−126 cos X1sin X2wv. = 0

For a fraction to equal zero, the numerator must equal zero.

All that garbage in front of cos X is a constant. Thisimplies cos X = 0whichimpliesX{|*G*{}~ = 90°�270° • By considering the original figure, only the 90° answer makes sense.

Notice this method does still get the correct result but using a little physics reasoning is a lot less tedious.

7

2.3 Figure is drawn for you. This time we get the following

table of knowns and unknowns. Consider the big 3 constant �� kinematics equations:

�) = �* + �*+ + 12 �+. �)+ = �*+ + �+ �)+. = �*+. + 2�+1∆�2 Notice in the third equation the only unknown is �*+!

First plug in the zeros and solve for �*+: �*+ = J−2�+1∆�2 Now plug in the knowns:

�*+ = N−21−# sin X2 Z34 V[ = N32#V sin X

Notice the minus signs under the square root work out. Check the units: m��( ∙ m = �� . Looks good.

Recall # = +9.8 ��(. The acceleration can be ± but # is the magnitude of the acceleration (due to gravity during freefall).

# is always positive…

ALWAYS POSITIVE

The acceleration can be ± but # is the

magnitude of the acceleration (due to gravity

during freefall).

Can you tell I’ve had to say this literally thousands of times over my career…

�* 0 (assume origin at start)

�) 34 V

�*+ ? �)+ 0 (in 1D motion an object comes to rest momentarily at max/min) �+ −# sin X (negative when arrow points opposite coordinate system) ?

8

2.4 By the way, # is always positive.

Part a: I would go straight to the velocity equation �)+ = �*+ + �+ �)+ = 3.834ms + 5−4.9 ms.6 11.00s2 �)+ = −1. 066ms

I wrote 4 sig figs but would round to 2 (see underbar). If I

need to use this number later, I want subsequent answers

to be good to 3 sig figs unless otherwise specified.

Part b: If you simply plug in the numbers from the problem you get a displacement ∆� = 1.38m. This is not

correct. In 1D motion displacement does not equal distance traveled when the object changes direction. It started

with positive velocity (going with the positive direction of the coordinates) and after one second we found a

negative velocity (going against the coordinates). One procedure is to do as follows:

• First find the time to reach zero velocity (this is the turnaround time).

• Determine the time for the second stage (going back downhill) by subtracting this time from 1.00 sec.

• Find the displacement on the way up using the first time.

• Find the displacement on the way down using the second time. Note: this should be a negative number.

• Take the absolute value of each displacement and add them together for the total distance.

�)+F = �*+F + �+FF → 0 = �*+F − #2 F

This leads to

F = 2�*+F# = 2 53.834ms 65+9.8 ms.6 = 0.7824s �)+F. = �*+F. + 2�+F1∆�F2 → 0 = �*+F. − #∆�

This leads to

∆�F = �*+F.# = 53.834ms 6.5+9.8 ms.6 = 1. 500m

. = 1.00sec − F = 0.2176s

∆�. = �*+.. + 12�+... →∆�. = −#4 ..

Plugging in numbers gives

∆�. = −5+9.8 ms.64 �0.2176s�. = −0.1160m

The negative sign makes sense because it is displaced

down the ramp after reaching max height.

�O��!�������� = |∆�P| $ |∆�Q| = P. SPQ�

�O��!��� !���"��� = ∆�P + ∆�Q = P. T���

Notice the total length of the ramp is irrelevant. In real life we often have more information than is needed to solve

problems. One skill I hope you learn is how to determine if information is irrelevant to the problem at hand.

Entire 1.00 sec of travel Δ� ? (Ramp is 2.00 m doesn’t mean block slides 2.00 m…) �*+ 3.834ms

�)+ ?

�+ −# sin X = − o. = − F. 5+9.8 ��(6 − 4.9 ��(

(neg cuz arrow points opposite coord. sys.) 1.00s

Stage 1 (Up to Max Height) Stage 2 (After Max Height) ∆�F ? (Just because ramp is 2 m doesn’t mean block slides 2 m…) ∆�. ? �*+F 3.834 m/s �*+. 0 (in 1D motion object @ rest momentarily at max/min) �)+F 0 (in 1D motion object @ rest momentarily at max/min) �)+. ? �+F −# sin X = − o. (neg cuz arrow points opposite coord. sys.) �+. −# sin X = − o. (neg cuz arrow points opposite coord. sys.) F ? . 1.00sec − F

9

2.5

a) We need to do two separate problems. I like to split the page vertically like this:

�)+F. = �*+F. $ 2�+F1∆�F2 → �)+F. = 2�+FV

To keep it neat I chose to plug in �+F at the end. �)+F = ±J2�+FV

Down the plane is + so use positive root! �)+F = J2�+FV

Use this in � equation giving �)+F = �*+F + �+FF → J2�+FV = 0 + �+FF J2�+FV = �+FF

F = 1�+F J2�+FV

F = N2�+FV�+F.

Solving for F gives

F = N2V�+F = N 14V5# sin X

To solve for F the work should be nearly identical. All

the subscripts change to 2’s instead of 1’s. This is one

example of the benefits of doing problems algebraically.

We find

. = N2V�+. = N 10V3# sin X

Units check out.

As V increases, time increases as expected.

As X increases, time decreases as expected.

The shell had a smaller � and has smaller . Looks good.

For the two to finish at the same time, the time delay must equal the time difference.

∆ = . − F = N 10V3# sin X − N 14V5# sin X = kN103 − N145 lN V# sin X = 0.1524N V# sin X

b) I made a table using V = 2.00m and # = 9.8 ��(. The calculations for 4.0° are in bold.

c) It’s pretty easy to measure 2.00m with minimal error so the length is not a problem as far as errors are

concerned. The smaller the angle the longer the times. Reaction time will cause negligible errors for the

small angles. Unfortunately, for extremely small angles the bowing of the track is no longer negligible.

Also, reading an angle indicator in a class demo is usually ±0.5° if we’re lucky…more likely ±1°. Percentagewise, this error is absurd (1°outof4°is25%‼!2. For small angles the angle measurement

gives you headaches, for large angles the time measurement becomes problematic. Probably best to use a

larger angle and remove the human element by capturing a video with an electronic release…but that can

get blurry at high speeds...

d) At 90° the objects are in freefall (instead of rolling down the hill)! They hit the ground at the same time!

e) The problem mentioned rolling without slipping. If you make the angle large enough the balls will be both

rotating and sliding (rolling with slipping). The accelerations given to you are not valid in that case! Real

life is complicated. Most physics models make a lot of assumptions. Turns out, your models are worthless

if you create a scenario that violates your initial assumptions!

Solid Sphere Spherical Shell ∆�F V ∆�. V �*+F 0 �*+. 0 �)+F ? �)+. ? �+F

57 # sin X (+ cuz arrow points with coord. sys.) �+. 35 # sin X (+ cuz arrow points with coord. sys.) F ? . ?

10

2.6 Worst case scenario (minimum rate of slowing) the car stops at the last possible moment before hitting the deer.

∆�F = �| �)+.. = �*+.. + 2�+.1∆�.2 → 0 = �. − 2�∆�

This leads to

∆�. = �.2�

It is worth repeating � is the magnitude of �� (�� = −��̂).

We also know a connection between the two parts. C = ∆�F + ∆�.

No absolute values are required on the ∆�’s because the direction of motion is always positive for this problem.

Combining the stages with the connecting equation gives

C = �| + �.2�

Solving for � gives

� = �.21C − �|2 The units check. We expect � should increase if you take longer to react; the equation reflects this because as |

goes up the denominator decreases producing a larger value for �. Notice the opposite is true if C or � increases

which is also as we expect. As the problems get nastier, having ways to check and think about your results will

hopefully decrease mistakes and increase your odds of passing this course…

Reality check: Suppose you are 1/10 of a mile 1≈ 160mor530ft) from a deer travelling 60mph = 27�� D30�� . Your reaction time plus time to foot on your brake is perhaps 0.5sec. These combine to give

� = 530ms 6.160m − 30ms 10.5sec) = 6.2 ms. D 23#

I read some info online from Car and Driver which indicated performance vehicles can brake at a rate of about 1#.

For an average car then, hopefully this is about as close to the deer as you could get without hitting it.

Stage 1 (while reacting) Stage 2 (while braking) ∆�F ? ∆�. ? �*+F � �*+. � �)+F � �)+. 0

�+F 0 (while reacting brakes are not applied) �+. −� (neg cuz assuming positive to the right and � is

magnitude of acceleration (rate of slowing)) F | . ?

� � � = 0

11

2.7

Here we know the final speed is 75% less than the initial speed.

Said another way, the final is is 25% of the initial speed.

Said another way, �) = 0.25�* = 0.25�.

We also know the distance in this problem. Once again I would reach for the trusty �. equation. �). = �*. + 2�Δ� 10.25�). = �. + 2�C 0.0625�. = �. + 2�C −0.9375�. = 2�C

� = −0.9375 �.2C

� = −0.469 �.C

Here the negative sign indicates the acceleration is to the left. This makes sense as the bullet is moving to the right

and should be slowed down by contact with the board. For fun, imagine the board is 1-inch thick plywood and the

bullet is moving with initial speed 250 m/s (subsonic). Since 1 inch is 2.54 cm, half that is about 0.0254 m. I found

� = −0.469 5250ms 6.

10.0254) D 1.153 × 10�ms.

Note: if you assume the entire length of the bullet must travel through the board…we must add in the length of the

bullet to the distance traveled. If we say the bullet is perhaps 2.00 cm long this increases the distance to 0.0454 m

and decreases the acceleration to 6.453 × 10� ��(.

You can tell by these insanely large numbers my assumptions are probably more than a little off. I suspect the

bullet’s speed is typically reduced by only 10% or something for this thickness of board…

12

2.8 Dropping the rock is pretty much freefall from short buildings. Running a simulation I found online, dropping a

baseball from the top of a 20m D 70ft building changed the time of fall by about 5%. For taller buildings freefall

will work as long as you are willing to accept 5-10% error. Both the plane and the rocket experience forces other

thang gravity so they are not in freefall.

The moon orbiting the earth has only gravity acting on it. Since the sun and earth are both exerting gravitational

forces on it you could argue it is a little complicated. Note: the sun exerts about twice as much gravitational force

on the moon as the earth does. Still gravity is the only force acting on it and if you want you could call that freefall.

Unfortunately, you cannot use the constant acceleration equations of motions to compute anything for the moon. If

you are interested in astro you might read about the Roche-Hill sphere which discusses when a satellite (such as the

moon) can be considered as orbiting a planet as opposed to the sun.

The figure at right is supposed to show Superwoman on top of a mega-mountain throwing

rocks with different speeds (not to scale). The mega-mountain is so tall that the peak is

above the atmosphere so air resistance is negligible. Notice the rocks fall to earth.

As the rocks are thrown faster and faster, the curvature of the earth extends the time of the

fall to the surface more and more. If thrown fast enough, the surface of the earth curves

away so much that the rock is able to orbit the earth! This is supposed to show that orbiting

objects are essentially in in freefall, falling towards the earth, they just never get to the

surface. As they orbit they keep falling, and falling, and falling…

2.9 Your own reflections.

2.10

a) There is more weight pulling on the medicine ball but also more mass to accelerate. They should hit the

ground at the same time. This assumes air resistance is negligible, a reasonable assumption for short drops.

HOWEVER, for a large drop we expect air resistance to be non-negligible…

b) Air resistance makes less than 5% difference according to simulations. Most of the time the speed is less

than 10 mph. If you drive at 10 or 15 mph and stick your hand out the window, you don’t feel insane

amounts of drag. It seems reasonable to use freefall as a model.

c) If we assume they are the same radius, it is tempting to think they have the same area and thus the same

amount of drag force. From experience you probably know dense objects will cut through the air a little

better and reach slightly higher speeds (e.g. drop a leaf vs drop a rock). From experience, we expect air

resistance increases with increasing speed. The more dense of the two (the medicine ball) will actually go a

little faster and thus have a little more drag force. It also has a lot more weight. The more dense object has

more drag force while also winning the race. The drag force per unit mass is less on the medicine ball but

the total drag force is greater.

2.11 & 2.12 Look at my supplemental handout on robjorstad.com. Use the simulation! Air resistance changed the

time by no more than 5% for all the different pre-set objects. You can try it out with the user defined and go nuts if

you want.

13

2.13

a) One way to draw the figure is shown at right. I tried to think

about all the important positions in the problem. For me,

initial position, max height, and impact position jumped out as

important.

b) To max height we know the list

of knowns shown at right. For

max height problems, I

typically consider the �. first as

it often helps. �F)�. = �F*�. + 2��Δ?F �F*�. = −2��Δ?

�F*� = ±m−2��Δ?

�F*� = ±J−21−#)Δ? �F*� = ±J2#Δ?

�F*� = ±m259.8 ms.6 14.00m) �F*� = 8. 854ms

I used positive root because in picture I can see it is initially moving upwards. To clarify this is a vector

one might write ��~}��{� = 8. 854�� �̂ c) You could use the Δ? equation…but the � equation is easier. �F)� = �F*� + ��F

F = −�F*��� = −58. 854ms 65−9.8 ms.6 = 0.9035s d) Since I know the time for the first stage already, I might as well get the time

for the second stage and add them together. For stage 2 (max height to

impact) we find

Δ?. = 12��.. + �.*�.

Δ?. = 12 1−#)..

. = N−2Δ?.# = N−21−6.00m)9.8 ms. = 1. 1065s The total time is thus GHG = F + . = 2. 01s To get the final sig figs I did not use the number of sig figs. I kept the left most column when adding.

Alternatively, for the entire flight we know the total displacement is Δ?GHG}~ = ?.) − ?F* = −2.00m. Plug

this into Δ? = F. ��GHG. + �*�GHG and do the quadratic formula. This is a great method as well.

e) NO. Total flight time is twice time to max height IF launch position and impact position at same height.

f) You could go from initial to impact and use GHG or max height to impact using .. Find �.)� = −10.84�� .

To clarify this is a vector, we might write ��*�I}{G = −10.84�� �.̂ Problem continues next page…

∆?F 4.00m �F*� ? �F)� 0 �� −# = −59.8 ��(6 F ?

∆?. −6.00m

Neg cuz went down �.*� 0 �.)� ? �� −# = −59.8 ��(6 . ?

�F)� = �.*� = 0

�1�? =?

�2�? =?

?F* = 2.00m

?1� = ?2� = 6.00m

?2� = 0.00m

14

g) To get distance traveled we must split into stages. Distance is not equal to displacement when the ball

reverses direction! It takes 0.9035s to the top and travels 4.00m upwards. In the next 0.3465s the ball is

displaced downwards from max height −0.5883m. Total distance traveled is �∆�GH�}+��*o�G� + |∆�@.�@v�GHF..��| = 4.588m

h) Displacement after 1.25s is much easier. You could just stick 1.25s into the Δ? equation. Since I already

have the numbers from the stages it is faster to use those. The displacement is 3.412m. To clarify this is a

vector we might choose to write this as 3.412m�.̂ i) The position of the ball after 1.25s can be found using Δ? = ?) − ?* ?) = ?* + Δ? ?) = 2.00m + 3.412m ?) = 5.412m

Recall, there is an implied �̂ on this result. This is a position vector, not a scalar distance. The positive

nature of the vector indicates the position is 5.412m above the origin. In this case the origin is at ground

level.

j) Total distance can be found using numbers in the original problem statement (or picture). 10.0 m.

k) Total displacement is -2.00 m.

l) Final position is 0.00 m.

m) The impact speed (no vector…just the magnitude…don’t need minus sign) is about

10.84ms × 1mi1609m × 3600s1hr D 24mph

A quick trick is to multiply by about 2 (or 2.2) to approximately convert m/s to mph. this is handy for

comparing your physics work (typically in m/s) to speeds you may be more familiar with (typically in

mph). Air resistance probably plays a significant role but our answers should still be in the right ballpark.

15

2.14

a) Unless otherwise noted, we assume ��is aligned with the positive direction of the coordinate system. In this

case, that implies �� = ���̂. This means �� = −# where # = +9.8 ��(. The first equation is not correct as

you should not explicitly write the negative sign on ��. In the third case, you should explicitly write the

negative sign once you have subbed in �� = −#.

b) Givens are listed in the table at right. After writing down the big three kinematics

equations and plugging in the knowns we find

∆? = �*� + 12 ��. →−ℎ = −#2 .

= N2ℎ#

Check the units: all good. As ℎ increases, increases as we expect. On the moon, where # is less, the fall

would take longer. Looks good.

c) You could plug in the time from the above equation into

�)� = �*� + �� → �)� = 1−#)kN2ℎ# l = −J2#ℎ

By using the �. equation I can start fresh from the givens. If for some reason I made a mistake in

computing my error will not propagate. Check with the �. equation gives �)�. = �*�. + 2��1∆?) → �)�. = 21−#)1−ℎ) = ±J2#ℎ

Notice in the problem we expect the ball is moving downwards just before impact. According to the

coordinate system chosen we expect �)� < 0 which indicates we should choose the negative root!

Check the units: all good. As ℎ increases, � increases as expected. On the moon # is less, the final speed

is slower. As I just mentioned, �)� < 0 as expected. Looks good.

d) The impact velocity is ��) = −J2#ℎ�̂. The impact speed is ���)� = J2#ℎ. Carefully look at the wording

on test questions to see if the instructor asks for speed or velocity.

Note: common convention in 1D motion is to leave the �’̂s and �̂’s unwritten and assume positive numbers

are to the right and up while negative numbers are to the left and down respectively. Using this convention

the final velocity is �) = −J2#ℎ while the final speed is �) = J2#ℎ. I know it is pretty annoying, but it

seems to be common among physicists. To make your final answers unambiguous to your instructor,

consider using the �’̂s and �̂’s for velocities.

Solution continues on next page…

∆? −ℎ �*� 0 �)� ? �� −# ?

16

e) For travelling half the time to the ground use the list of knowns at right.

Notice I took the time from part b and divided by 2.

Using the ∆? equation gives

∆? = �*� + 12 ��.

∆? = −#2 .

∆? = −#2 kN ℎ2#l.

∆? = −ℎ4

Again we should think about what the minus sign means.

The displacement vector is ∆?� = − �� �̂. The distance traveled is ‖∆?�‖ = ��.

Also, you should find �)� = �*� + �� �)� = 0 + 1−#) �)� = −#N ℎ2#

�)� = −N#. ∙ ℎ2#

�)� = −N#ℎ2

This gives speed ���)� = mo�. which is half the impact speed 5F. ofJ2#ℎ6.

f) For the time to travel half the distance use the set of givens shown at right.

∆? = �*� + 12 ��.

− ℎ2 = 0 − #2 .

�(��� ¡¢£¤ = Nℎ#

Notice, if we look back at the TOTAL time from part b

�(��� ¡¢£¤ = GHG}~√2 D 70%ofGHG}~ You should find ���)� = J#ℎ which is about 70% of the result for the total fall!

Speed after falling half the distance is not half of impact speed!

∆? ? �*� 0 �)� ? �� −#

12N2ℎ# = N14 ∙ 2ℎ# = N ℎ2#

∆? −ℎ/2 �*� 0 �)� ? �� −# ?

17

2.15

a) From ground to max height (ignoring Jackie’s height) gives the

figure and list of knowns at right. For max height, I love to attack

using the �. equation: �)�. = �*�. + 2��1∆?) 0 = �. + 21−#)∆? −�. = −2#∆? −�. = −2#∆?

∆? = �.2#

A couple of points worth noting.

• 1−�). = �. while −�. = −1�.)… notice the subtle difference

• Remember # is always positive. This can help you catch sign errors.

• This equation tells you how far the ball travels above its initial position. If the ball had started on

top of a cliff this still works but this gives the max height above the cliff, not the ground.

• At max height, �)� = 0. For balls thrown at angles (Chapter 4), the speed at max height is non-

zero if �*+ ¥ 0. In general, at max height �� = �*+�̂ + 0�̂. More on this later.

b) To max height we continue to use the same picture and list of knowns as in part a. From the � equation �)� = �*� + �� 0 = � − # �}+ ��*o�G = �#

c) For the entire flight the picture and unknowns change to those seen at right.

Using the ∆? equation gives

∆? = �*� + 12 ��. → 0 = � − #2 .

Solving for gives

GHG}~ = 2�# = 2�}+ ��*o�G

d) Use the � equations to show

�)� = � − # = � − # Z2�# [ = −�

Notice the final velocity is ��) = −��̂ while the final speed is ���)� = �.

What goes up, must come down at the same speed…if air resistance and initial height are negligible.

The acceleration is constant! We are using constant acceleration equations of motion. The acceleration at

max height is the same as the acceleration everywhere else �� = −#�̂ = #1−�̂) If you answered ¦ you are wrong.

Note: the problem asked for acceleration, not the magnitude of acceleration.

I’m not trying to be a jerk; proper statements are critical in this field.

Note: an alternate correct way to express this result is �� = −#.

∆? ? �*� � �)� 0 �� −# ?

∆? 0 �*� � �)� ? �� −# ?

#

�

�� = 0

#

�

18

2.16

a) The rock thrown upwards has a picture and list of givens shown at right. The

impact velocity is found using �)�. = �*�. + 2��1∆?) → �F)�. = �. + 21−#)1−ℎ) �F)� = ±J�. + 2#ℎ = −J�. + 2#ℎ

Notice in the problem the rock is moving downwards just before impact.

According to the coordinate system chosen, we should choose the negative root!

The impact velocity is ��F) = −J�. + 2#ℎ�.̂ The impact speed is ���F)� = J�. + 2#ℎ.

b) Use this �)� in the � equations which gives �)� = �*� + �� −J�. + 21#)1ℎ) = � − #F

�P = § + J§Q + Q¦¨¦

This result is acceptable. Physicists often try to rewrite answers such that they are comprised of a term

with correct units times a term with no units. This rearrangement is shown below.

F = � + m�. + �.�. ∙ 2#ℎ#

F = � + m�. Z1 + 2#ℎ�. [#

F = � + �mZ1 + 2#ℎ�. [#

�P = §¦ ©P + NP + Q¦¨§Q ª Notice the second bold result has the correct units outside the brackets and no units inside the brackets.

Notice, however, this form of the answer makes no sense when � = 0. Either bold result is fine.

c) If we plug in � = 0, we find �F)� = −J2#ℎ and F = m.�o just as we found in the previous problem.

Solution continues on the next page…

∆?F −ℎ �F*� � �F)� ? �� −# F ?

#�

19

d) The rock thrown downwards has a picture and list of givens shown at right. The

impact velocity is found using �)�. = �*�. + 2��1∆?) �.)�. = �. + 21−#)1−ℎ) �.)� = ±J1−�). + 2#ℎ �.)� = −J�. + 2#ℎ

The coordinate system and impact direction imply use the negative root.

The impact velocity is ��.) = −J�. + 2#ℎ�.̂ The impact speed is ���.)� = J�. + 2#ℎ.

Notice this is exactly the same speed as the rock thrown upwards! This makes sense because we know

from problem 2.13d the ball thrown upwards should, in the absence of air resistance, come back to its

initial height with identical speed moving in the opposite direction. From that point on, the problems are

identical and should reach the same final speed.

Use this �)� in the � equations which gives �)� = �*� + �� −J�. + 21#)1ℎ) = −� − #.

. = J�. + 2#ℎ − �# = �# ©N1 + 2#ℎ�. − 1ª Notice the importance of getting signs correct while listing your givens and during your algebra!

e) Since F is longer (rock thrown upwards), the time difference is given by ∆ = F − .

Δ = �# ©1 + N1 + 2#ℎ�. ª − �# ©N1 + 2#ℎ�. − 1ª Δ = 2�#

Think! Why does this make perfect sense? That is the extra time for the ball to go to max height and back

to the initial position…just like problem 2.13c!

f) The distance to max height is given by ∆? = «(.o as in problem 2.13a. The total distance traveled is ��C�¬�A­� = 21C�¬�A­���� ℎ��#ℎ) + ℎ

��C�¬�A­� = 2®�.2#¯ + ℎ

�O��!�������� = §Q¦ + ¨

∆?. −ℎ �.*� −� �.)� ? �� −# . ?

#�

20

2.17

a) The ring released from the hand travels downwards with the same speed as Xtremo.

According to the coordinate system chosen, anything going downwards now gets a plus

sign while anything going upwards gets a minus sign. While at first this seems weird,

notice we now have only positive signs in the list of givens. Hopefully this will cause

fewer minus sign errors.

Using the �. equation gives �)�. = �*�. + 2��1∆?) → �)�. = �°. + 21#)1ℎ) �)� = ±m�°. + 21#)1ℎ) = m�°. + 2#ℎ

Notice in the problem we expect the ring is moving downwards just before impact. According to the

coordinate system chosen, this indicates we should choose the positive root!

b) For the ring we can use this �)� in the � equations which gives

�)� = �*� + �� → m�°. + 21#)1ℎ) = �° + #°

° = J�°. + 2#ℎ − �°#

The time for Xtremo to hit the ground is much simpler. Since acceleration is zero for the constant speed,

we may use C�¬�A­� = 1���) × 1���) giving ± = t*�G}�{�|}G� = �«².

The difference in time between the two is thus³�G´��� = ± − ° = ℎ�° − J�°

. + 2#ℎ − �°#

c) We expect as ℎ → 0 that ³�G´��� = 0 if we ignore the height of the Xtremo. Plugging in ℎ = 0 and

simplifying gives this result. Doing these kind of checks can help you find your own mistakes on a test.

Extra Solutions for Part b

Maybe you noticed this is the same thing as solving � = ∆«∆G to get ∆ = ∆«} . That’s cool.

Another, more cumbersome way to solve this, is to use the ∆? equation.

∆? = �*� + 12 ��. → ℎ = �° + #2 .

Rearranging gives #2 . + �° − ℎ = 0

Doing the quadratic formula for using � = o., µ = �° , and ­ = −ℎ gives

� = −�° ± m�°. − 45#26 1−ℎ)2 5#26 = = ±J�°. + 2#ℎ − �°#

To figure out the root you must compare the term in the square root to �°. By factoring out �° it is easier to

see the square root must use the positive root to make time positive. It is also easier to check the units.

Doing a symbolic quadratic formula is good practice. They get easier. I especially enjoy the part about

trying to figure out when to use the ± sign for the root. That, to me, is the fun part.

= ±«²NFw(¶·¸²( Y«²o = «²o Z±m1 + .o�«²( − 1[ = «²o Zm1 + .o�«²( − 1[

∆? +ℎ �*� +�° �)� ? �� +# ?

21

2.18 The picture would like the one shown at right.

Note, we know for a level ground problem (the first half

of the flight), the ball will have final vertical velocity

equal in magnitude but opposite in direction. This is the

trick I used to know �1�? = �2�? = −�.

The time to go up and back to the same vertical position

(from ?F* to ?F)) can be found using

Δ?F = �F*� + 12 ��.

0 = � − 12#.

= 2�#

The problem tells us this is half of total flight time.

Therefore, for this special case, this time also equals the

time to go from ?.* to impact!

Δ?. = �.*� + 12 ��.

−ℎ = 1−�) + 12 1−#).

−ℎ = 1−�) 2�# + 12 1−#) Z2�# [.

Cancel all minus signs and clean up some terms

ℎ = 2�.# + 2�.#

¨ = �§Q¦

�1�? = �

�1�? = �2�? = −�

?F* = ℎ

?2� = 0.00m

?F) = ?.* = ℎ

22

2.19

a) We are assuming the acceleration time for both the rabbit and the

tortoise are negligible. Note: we do not need to know the exact

point in time when the hare takes a break. If the tortoise runs for

time we do know that the hare will run for − ³ regardless of

when the break is taken!

Solve the tortoise equation for v and plug it into the hare equations.

This will eliminate the unknown v so you can solve for the

unknown t. You should find

C − V = 20C 1 − ³) C − V = 20C − 20C ³

= 20C19C + V ³ = 20118.9m)19118.9m) + 0.900m 1120s) = 126s From here use �}|� = − ³ = 6s.

b) The speed of the tortoise is straightforward

� = C = 18.9m126s 1mi1609m3600s1hr = 0.3356mihr = 0.336mihr

For the hare you should find ��}|� = 20.0� = 6.71�p¹º. It is worthwhile to double check this result using

the distance traveled by the hare over the time traveled by the hare. Consider the equation below and make

sure you understand it.

� = C − V − ³ = 18.0m6s 1mi1609m3600s1hr = 6. 71mihr

Notice that when doing the problem this way we would have only one sig fig if following sig fig rules…

List of knowns

C = totallengthofrace = 18.9m

V = lengthofshell = 0.900m

³ = ³|�}¼ = 2.00min = 120sec ��}|� = 20.0�

Equation

for tortoise

Equation

for hare C = � C − V = 20�1 − ³)

23

2.20 This is an interesting problem. When you first read it, it seems ambiguously worded.

When it reaches the person, couldn’t the ball be moving either upwards or downwards?

To me, the phrase “takes times to reach the person” implies the ball first reaches the person in time . This implies the ball must still be on the way up…

Δ? = �*� + 12 ��.

ℎ = �*� − 12#.

�*� = ℎ + 12#.

§�½ = �̈ + ¦�Q

24

2.21 For max height problems, I usually try the �. equation first…

WATCH OUT! The max height ?�}+ is unknown. If you go from the initial

launch position to �¾¡¿� (see figure at right �)�. = �*�. $ 2��Δ?

�. = �*�. $ 210#2 ?�}+4

�*� = N�. $ #�2

This answer includes the unknown ?�}+!!! This is not yet solved!!!

Comparing launch position to max height might get you there…but first I will

try comparing �¾¡¿� to ?�}+. �)�. = �*�. $ 2��Δ?

0 � �. $ 210#2 5?�}+ 0 ?�}+4 6

0�. � 02# Z34 ?�}+[

HEY YOU! WATCH OUT! 0§Q � 01§Q2 The minus sign is not canceled by the squared here. It is canceled by the

minus on the other side of the equation…

�. � 2# Z34 ?�}+[

�. � 32#?�}+

Hmmm… in this problem we are given �. This means we now know the max height in terms of the givens!

?�}+ � 2�.3#

Now compare launch position (? � 0) to ?�}+. Note: you could also compare launch position to �¾¡¿� but this is

produces slightly more algebra. �)�. � �*�. $ 2��Δ? 0 � �*�. $ 210#21?�}+ 0 02 �*�. � 2#?�}+ �*� � J2#?�}+

Here we know the ball is going upwards at the instant of launch. This is why I ignored the negative root.

�*� � N2# ®2�.3# ¯

�*� � N43� D 1.155�

Tip: sometimes it is prettier to write this as a decimal number, which is why I included both answers.

�?��� � 0

�2? � � ?�}+4

? � 0

?�}+

�1? �?

25

2.22 With this kind of problem students often wonder if the upwards acceleration already accounts for the effects of

gravity or if they need to subtract off # � 9.8 ��( from the upwards acceleration. The problem wording says “upward

acceleration with magnitude 2#”. This implies the effects of gravity are already included. Note: later we will learn

net force is given by À���G � ���. This implies the net force on the rocket is À���G � �12#2�̂. Since the weight pulls

down with magnitude �#, we know the upwards force from the thruster must be 3�#!

If the upwards acceleration is �� � 2# � 19.6 ��( we can use �)� � �*� $ �� to determine the rocket takes 5.0 s to

reach 98 m/s. At that point the rocket has traveled upwards 245 m (use the Δ? equation).

From this point it is a standard freefall problem with initial height 245 m, initial velocity 98 m/s upwards, and

acceleration now magnitude # directed downwards.

It will take 10.0 seconds to reach max height (use the �)� equation using �)� � 0 at max height).

In that time it travels upwards to an altitude of 735 m (use the �)�. equation �)� � 0 at max height).

At this point the rocket is now instantaneously at rest at max height. This height is 735 m above the ground.

Ignoring air resistance seems a bit ridiculous for that large of a distance, but we will ignore it anyways to get a rough

approximation of the flight.

It takes about 12.2 sec to hit the ground (use the use the Δ? equation with Δ? � 0735m and �*� � 0). Impact

velocity should be about -120 m/s. The total time should be about 27.2 sec.

0.0

100.0

200.0

300.0

400.0

500.0

600.0

700.0

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 22.0 24.0 26.0 28.0

x (m)

t (s)

-150.0

-100.0

-50.0

0.0

50.0

100.0

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 22.0 24.0 26.0 28.0

v (m/s)

t (s)

-20.0

-10.0

0.0

10.0

20.0

0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 22.0 24.0 26.0 28.0

a (m/s2)

t (s)

26

2.23 I choose to call down the positive direction. While thruster is on

magnitude of acceleration is ��# directed upwards (negative direction).

The moon’s gravitational force cancels F� of the thrust!

Thruster OFF Thruster ON

∆?F �F �? ∆?.

�. �? 1butwedoknow�. � à 0 �F2 Assumes à � 200m

(Positive because down is positive direction) �*�F � � 20ms �*�. ? 1butwedoknow�)�F � �*�.2 �)�F ? �)�. 0 1Iflandercomestorestexactlyatthesurface2

��F $ F�# ��. 056#

(Negative because down is positive direction) F ? . ? We don’t know any of the times but we do know about at least one velocity and some relationship with the heights.

This suggests to me I might write down the �). equation first to see if I can get anywhere with that.

Thruster OFF Thruster ON �)�. � �*�. $ 2��1∆?2 �F)�. � �. $ #3 �F

�)�. � �*�. $ 2��1∆?2 0 � �F)�. 0 53#�. Note:weknow�)�F � �*�..

Plugging in from Thruster OFF equations gives

0 � 5�. $ #3 �F6 0 53#�.

Now use �. � Ã 0 �F to reduce # of unknowns

0 � �. $ #3 �F 0 53#1Ã 0 �F2 0 � �. 0 53#Ã $ 2#�F

�F � 56Ã 0 �.2#

�F � 56 �200� 0 20. 0.2�9.80� �F � 146.26m

Note: I assumed 3 sig figs on all givens but notice the

final answer only has 2. Keeping the rest of sig figs to

reduce errors when checking the solutions.

Once you know one piece of the puzzle the rest all works out. You should find �. � 53.74m. Depending on the

method used to solve the sig figs might vary.

If you choose to subtract directly from 200m, you end up with only one sig fig.

If you use 0 � Z�. $ ov 1Ã 0 �.2[ 0 �v#�. � �. $ Fv#Ã 0 2#�., you probably get two sig figs (I didn’t check).

Final answer: about 54m above the lunar surface is the position to fire the thrusters.

à �200m �F

�2

16#

56 #

27

2.24 I assumed up is positive direction.

Kinematics from top to bottom of window gives

0� � �°F 0 #F.2

�°F � 0� $ #F.2

�° � 0 �F $ #F2

Also we know �È � 0�° 0 #F

�È � 0 �F 0 #F2

From bottom of window to impact gives �) � �È 0 #.

�) � Z0 �F 0 #F2 [ 0 #.

�) � 0 �F 0 # ZF2 $ .[

Kinematics on the entire fall (top of building to impact) gives �). � �*. $ 210#210Ã2 �) � 0J2#Ã

Answer for §É is negative because traveling down at impact.

Comparing these results gives

à � 12# Ê�F $ # ZF2 $ .[Ë.

There are other ways to write this as well. Consider # � 10 ��(, � � 2m, F � 0.4s, and . � 2s. I found à D 36.5m. You should find the top of the window is 1.25m

below the roof (or bottom of window is about 33m above the ground).

Clearly my figure is not to scale.

�°

�È

�)

Ã

�, F

�* � 0

. from

bottom

of

window

to impact

28

2.25 I ignored sig figs here to focus on concepts, units, and other things.

a) ��* � 06m� ̂and ��* � 04�� � ̂b) To get a decent estimate, use a range of times with the time of interest (3 s) somewhere in the middle.

I drew the line tangent to the curve at 3 s. I made a little triangle on my paper. I did rise over run. ��¬��BA D ��) 0 ��*) 0 * � 15m�2̂ 0 105m�2̂13.5s2 0 12.3s2 � 8.3ms � ̂Your number is probably a little different from mine but hopefully within about 10%. Comparing to the � graph, it appears to be in good agreement.

c) The area under the � curve can be calculated using

Ì��� � 12 µ�¬� > ���#� � 12 11.0s2 > 504ms �̂6� 02m� ̂Notice this is not the position at � 1.0s but the

displacement between � 0to1.0s. d) From the intercept of the � graph we known initial

position is ��* � 06m�.̂ From part c we know ∆�� � 02m�̂. We can rearrange ∆�� � ��) 0 ��* to give ��) � ��* $∆�� � 08m�̂ which agrees with the � graph.

e) For � 1.0to4.0s we find ∆�� � 18m�.̂ Ì��� � 12 µ�¬� > ���#� � 12 13.0s2 > 512ms �̂6 � 18m� ̂

WATCH OUT! The base of the triangle is 14.0s 0 1.0s2 � 3.0s. f) The total displacement for � 0to4.0s is ∆�� � ∆��@GHF $ ∆��FGH� � 02m�̂ $ 18m�̂ � 16m� ̂

Notice on the � graph the initial position is ��* � 06m� ̂and the final position is ��) � 10m� ̂which gives

the same total displacement.

g) Total distance is different. C�¬�A­� � ‖∆��@GHF‖ $ ‖∆��FGH�‖ � ‖02m�‖̂ $ ‖18m�‖̂ � 20m

Notice displacement is a vector while distance is a scalar (no �)̂. Both have units of meters…

h) The total area under the � graph is 14.0s2 > 54.0 ��( �6̂ � 16�� �.̂ This total area is CHANGE in velocity.

The initial velocity on the � graph is ��* � 04.0�� �.̂ The final velocity on the � graph is ��) � 12�� �̂. The change in velocity found using area under the � curve agrees with the numbers on the � graph.

i) I compute the slope at � 1.0s by considering a range of time with 1 s in the middle. ��¬��BA D ��) 0 ��*) 0 * �

54.0 ms �̂6 0 504.0 ms �̂612.0s2 0 10s2 � 4.0 ms. � ̂

Problem continues on next page…

29

j) If you are a math freak, you might be worried about the use of open versus closed intervals. While that is

admirable, I typically ignore that distinction. That said

2.26 Use the simulation. A suggested simulation is available for free in the supplemental handouts found on

robjorstad.com.

What I asked for What it implies Said another way When it occurs

Moving right,

speeding up �� > 0AND�� > 0

� is positive and slope

of �-plot also positive � 1.0to4.0s

Moving right,

slowing down �� > 0AND�� < 0

� is positive and slope

of �-plot is negative Doesn’t occur

Moving left,

speeding up �� < 0AND�� < 0

� is negative and slope

of �-plot also negative Doesn’t occur

Moving left,

slowing down �� < 0AND�� > 0

� is negative and slope

of �-plot is positive � 0to1.0s

At rest �� � 0 � is zero @ � 1.0s

30

2.27 A blow up of the �-plot between 1.5 and 2.5 s is shown at right.

a) Between 3.00 and 6.00 sec (�� and �� have same sign).

b) Between 1.95 & 2.00 s the particle goes from speed 4.00 m/s

to speed 0. It is slowing down.

c) Between 2.00 & 3.00 sec.

d) Between 0 & 1.95 sec and between 2.00 & 3.00 s.

e) I put all the plots on the next page. The position values can

be read off the upper plot (on the next page) or the bottom

right table (on this page).

Generally, we try to write three sig figs unless otherwise specified. I made an

exception in this case because I felt like it conveys the important aspects of problem

and reduces clutter in the results.

WATCH OUT! If a number has units, the numbers in your answers typically have

units. In this case, the units are in the column headings.

Notice the subtle difference between the table of displacements (upper table second

column) and the positions (lower table).

• Positions are tabulated at instants in time.

• Displacements are tabulated over time intervals.

• Displacements correspond to CHANGES in position.

Going further: Consider running a simulation of this scenario. To change a parameter

at one particular instant: pause the simulation momentarily, make the change, then

restart the simulation.

∆t (s) ∆x (m) a (m/s2)

0 to 1.0 4.0 0

1.00 to

1.95 3.8 D 4 0

1.95 to

2.00 0.1 D 0 -80.0

2.0 to

3.0 0 0

3.0 to

4.0 -1.0 -2.0

4.0 to

5.0 -3.0 -2.0

5.0 to

6.0 -5.0 -2.0

t (s) x (m)

0.0 -4.0

1.0 0.0

2.0 3.8

2.0 3.9

3.0 3.9

4.0 2.9

5.0 -0.1

6.0 -5.1

0

2

4

1.5 1.75 2 2.25 2.5

v (m/s)

t (s)

31

Plots for 2.27

Note: on the acceleration plot below the spike at about 2 seconds dips all the way down to 0�Ñ �UQ.

-5.00

-4.00

-3.00

-2.00

-1.00

0.00

1.00

2.00

3.00

4.00

0.00 1.00 2.00 3.00 4.00 5.00 6.00

x (m)

t (s)

-6

-4

-2

0

2

4

0 1 2 3 4 5 6

v (m/s)

t (s)

-8

-6

-4

-2

0

2

0 1 2 3 4 5 6

a (m/s2)

t (s)

32

2.28

a) Moving to the right implies positive velocity. This appears on an �-plot as regions of positive slope. Between

1.00 & 3.00 seconds AND between 5.00 & 6.00 sec the plot has positive slope and is thus moving to the right.

b) At 4.00 seconds we get a numerical value of velocity from the slope using

§�!O���½ � �!O �OÉ�� � Ò���ÒÓ� D 0P��Ñ. �ÑU � 0TÔ�U

I think anything in that 30-40 m/s range seems fine to me. When reading graphs you never really get 3 sig figs…

Note: I am assuming an implied � ̂is on this answer as stated in problem.

COMMON MISTAKE: You should pick points just before and just after the time of interest to ensure a good

estimate for the slope. I used the green points shown below to get the slope. The red points are a bad

approximation. The red points INCORRECTLY estimate the velocity as -20 m/s (off by more than 40%)!!!

c) Moving left implies negative velocity (neg slope on xt). Slowing down implies acceleration and velocity are

opposite signs. Here this implies negative velocity but positive acceleration. For an xt plot that means negative

slope but positive concavity. This occurs between 0.00 & 1.00 sec AND between 4.00 & 5.00 sec.

d) Moving right & speeding up implies positive slope & concave up xt plot. This occurs between 1.00 & 2.00 sec

AND between 5.00 & 6.00 sec.

e) Negative acceleration implies concave down. This occurs between 2.00 & 4.00 sec.

33

2.29

a) At rest implies velocity is zero. This occurs on a vt whenever the curve crosses the horizontal axis. Think: if the

plot crosses the vertical axis the vertical coordinate is zero. On a vt plot the vertical coordinate is velocity. I

estimated the times as 2.0 sec and 4.8 sec. Usually hard to get 3 sig figs from reading a graph…this part maybe ok.

b) At 2.00 seconds we get a numerical value of acceleration

from the slope. Normally you pick points, one just before

and one just after, to find the slope. In this case, because the

slope is constant, it is ok to use a larger time interval. I used

the two green dots shown in the figure.

�­­������A � ¬�Õ��� � ��¬��BA D 16.0 ms2.0s � 8.0 ms.

This might be an exception where you actually could get 3

sig figs from the plot…but I’ll still roll with two sig figs so

as not to express sig figs overconfidently.

c) Moving left and speeding up implies negative velocity and acceleration same sign as velocity. On a �-plot we

are thus looking for regions with negative values (negative vertical coordinates) with negative slopes. This occurs

between 4.8 & 6.0 sec.

d) Displacement is given by area under the �-curve

including negative signs for areas below the horizontal

axis. I see two obvious regions: from 4.0 to 4.8 sec and

from 4.8 to 5.0 seconds.

Ì@GH@.�Ö � 12µ�¬� > ���#� D 12 10.80s2 516.0ms 6� 6.4m

Ì@.�ÖGH..@ � 12µ�¬� > ���#� D 12 11.20s2 5024.0ms 6� 014.4m

Summing and keeping track of signs gives a total

displacement of about -8.0 m. Since the numbers line

up perfectly on the grid for this part, I’d expect your

numbers to be spot on.

e) The distance traveled is slightly trickier. Remember to do the same thing as the previous part but take the

absolute value of each section. Summing with absolute values gives a total distance of about 20.8 m.

×� � P. QÑU

Ø� = Ñ. �Ñ �

34

2.29¼

a) When looking at a �-plot, constant acceleration implies constant slope.

Said another way, constant a implies linear �-plot.

Objects 1 & 2.

b) When looking at a �-plot, positive acceleration implies positive slope (but slope can be changing).

Objects 1 & 3.

Note: One might argue the initial slope for object 3 is zero so, strictly speaking, it is not ALWAYS

increasing speed…just all times greater than = 0.

c) When looking at a �-plot, at rest means the value of � is zero.

Said another way, we are looking for the time or times when the graph intersects the horizontal time axis.

Object 4 is at rest at about 1.4 seconds. UNITS ARE REQUIRED ON ANSWER FOR FULL CREDIT.

Note: on exam days I try to be about 10% lenient when reading plots.

Any time between 1.3 s and 1.5 s would probably get full credit.

Note: when reading this graph, I argue it is pretty challenging to accurately get 3 sig figs.

I would be fine with two sig figs in this special case.

d) An object is speeding up whenever acceleration and velocity have the same sign.

On a �-plot, acceleration is the slope of the curve while velocity is the value of the curve.

We are looking for times when either:

• both the value of the curve and the slope are negative

• both the value of the curve and the slope are positive

Between = 0&1.4s the value of the curve is negative but slope is positive.

Between = 1.4&2.0s the value of the curve is positive & slope is positive.

Object is speeding up between = 1.4&2.0s. e) Speed and velocity are not the same thing!

Speed is the magnitude of velocity.

The final velocities are ��F) = +6.0�� �̂, ��.) = −6.0�� �,̂ ��v) = +4.0�� �̂&���) = −4.0�� �.̂ The final speeds are �F) = �.) = 6.0�� &�v) = ��) = 4.0�� .

The ranking is thus �F) = �.) > �v) = ��).

35

2.29 ½

a) When looking at a �-plot, initial velocity is the intercept of the curve with the vertical axis.

Initial speed is the magnitude of initial velocity.

The initial velocities are ��F* = +5.0�� �̂, ��.* = −2.5�� �̂, ��v* = +5.0�� �̂&���* = 7.5�� �.̂ The initial speeds are �F* = 5.0�� , �.* = 2.5�� , �v* = 5.0�� ,&��* = 7.5�� .

The ranking is thus ��* > �F* = �v) > �.* . b) On a �-plot, acceleration is the slope.

Objects 2 & 3 have constant slope (constant acceleration).

Object 1 has constant velocity. This implies acceleration is zero.

The question asked for non-zero constant acceleration.

c) Not moving implies velocity is zero. The best answer is “None of the objects” or “All objects move”.

I was trying to point out the distinction between not moving and not accelerating.

d) As discussed previously, object 1 has zero slope.

e) Reversing direction implies velocity changes sign.

On a �-plot, when velocity changes sign the curve must cross the horizontal time axis.

Only objects 2 & 3 reverse direction.

f) Final speed is the magnitude of final velocity. One finds �v) > ��) > �F) > �.).

g) On a �-plot, acceleration is the slope.

We are asked to find all objects with non-negative slope.

We already know object 1 as � = 0 and thus meets the requirements of the statement.

Object 2 also has non-negative slope.

Notice: asking for monotonically increasing velocity is slightly different than asking for positive velocity.

h) Monotonically changing speed means speed implies one of the following two conditions must be met

• Either the object never slows down.

• Or the object never speeds up.

Object 1 has constant speed and thus meets either requirement.

Object 1 is the only object that meets either requirement.

BE CAREFUL.

Object 2 slows from speed 2.50�� to 0�� (at 1.25s) then speeds back up to speed 3.75�� .

Object 3 slows from speed 5.00�� to 0�� (at 2.00s)then speeds back up to speed 15.0�� .

Object 4 speeds up from speed 5.00�� to 10.5�� (at 2.50s) then slows back down to speed 5.00�� .

i) On a �-plot, displacement is the area under the curve (areas below axis are negative, above are positive).

Object 1 displacement is +25.0m�.̂ UNITS ARE 1������¬BA�¬) × 1ℎ��Ú���¬BA�¬) = �� × s = m.

For object 2 between = 0&2.00s the displacement is F. 5−2.50�� 6 12.00s) = −2.50m�.̂

For object 2 between = 1.25&5.00s the displacement is F. 5+3.75�� 6 13.00s) = +5.625m�̂.

Total displacement of object 2 is +3.125m�̂. For object 3 between = 0&1.25s the displacement is

F. 5+5.00�� 6 11.25s) = 3.125m�̂. For object 3 between = 1.25&5.00s the displacement is

F. 5−15.0�� 6 13.75s) = −28. 125m�.̂ Total displacement of object 3 is −25.0m�.̂ Object 4 displacement is about 40m�.̂ I estimated the average height of the curve at 8.0�� .

The displacement rankings (most positive to most negative) are ��� > ��F > ��. > ��v.

j) To get distance, take the absolute value of each area before adding together.

The distance rankings are ‖Δ���‖ > ‖Δ��v‖ > ‖Δ��F‖ > ‖Δ��.‖.

36

2.29¾ WATCH OUT! Now we have a position versus time plot (instead of velocity versus time)!

a) On the �-plot, initial velocity means initial slope.

Initial speed means magnitude of initial slope.

The initial slope is about zero for objects 6 & 7.

The initial slope of 8 looks significantly larger than the initial slope of 5. �Û* > ��* > ��* = �Ö*On

b) On an �-plot, final velocity means final slope.

Object 6 has a steep positive slope.

Object 5 has a slightly less steep positive slope.

Objects 7 & 8 have negative slopes.

I zoomed way in on the graph and still had trouble noticing much difference between those two slopes.

I would say ���) > ���) > ��Ö) D ��Û)

c) On the �-plot, constant acceleration implies constant concavity.

The plots are all parabolic or linear.

They all have constant acceleration.

WATCH OUT! Object 5 is linear (which is neither concave up nor down).

Object 5 has zero acceleration!!!

Only objects 6, 7 & 8 have non-zero constant acceleration.

d) On an �-plot, non-positive acceleration means concave down (or linear).

Objects 7 & 8 meet this requirement.

e) On an �-plot, velocity is slope.

Speed is magnitude of slope.

Monotonically changing speed means speed implies one of the following two conditions must be met

• Either the object never slows down.

• Or the object never speeds up.

In other words, we are looking for slopes which never decrease or never increase.

Speed of object 5 never changes (meets requirement).

Speed of object 6 starts from rest and is ever increasing (meets requirement).

Speed of object 7 starts from rest and is ever decreasing (meets requirement).

Speed of object 8 is initially about 10�� then slows to zero at = 3.00s. Then object 8 reverses direction (slope changes sign) then speeds back up to about 8.0�� .

Object 8 does not meet the requirement.

f) Object 5 (as previously discussed).

g) Only object 8 as previous discussed.

h) An object at rest would have zero slope…it would look like a horizontal line.

i) Don’t over think it; on an �-plot, position is simply the value of the curve (don’t find slope or area).

Displacement is final position minus initial position.

Object 5 displacement is Δ��� = 115.0m)�̂ − 1−10.0m)�̂ = 25.0m�.̂ Object 6 displacement is Δ��� = 118.5m)�̂ − 1−12.5m)�̂ = 31.0m�.̂ Object 7 displacement is Δ��Ö = 1−20.0m)�̂ − 1−2.5m)�̂ = −17.5m�.̂ Object 8 displacement is Δ��Û = 17.5m)�̂ − 1−7.5m)�̂ = 15.0m�.̂ The displacement ranking (most positive to most negative) is Δ��� > Δ��� > Δ��Û > Δ��Ö.

j) Distance traveled requires some extra thought for any objects which reverse direction (object 8).

Split the displacement into chunks: Δ��Û)H|@→v� = +23.0m� ̂while Δ��Û)H|v→�� = −8.00m�̂. For objects with monotonically changing position, distance traveled is magnitude of displacement.

The distance ranking is therefore ‖Δ��Û‖ > ‖Δ���‖ > ‖Δ��Ö‖ > ‖Δ���‖.

37

2.29Ü� Watch out! I switched back to velocity versus time instead of position versus time. I’m hoping that going

back and forth between the two types will force you to learn the correct times to look for slope, are, and value of

curve (fingers crossed).

a) Take the absolute value of the initial value of curve on �-plot. �FF > �� > �F@ = �F.

b) Want constant non-zero slope of �-plot. Object 11.

c) Not moving implies � = 0 (horizontal line along time axis of �-plot).

All objects are move. None are non-moving.

d) Not accelerating implies � = ¬��# (horizontal line parallel to OR along time axis on �-plot).

Object 9 meets the requirement.

e) Reverse direction means velocity changes sign (crosses time axis of �-plot).

f) Object 11 reverses direction at = 2.50s. Objects 10 & 12 reverse direction at times D 0.50&4.0s.

g) Acceleration is slope of �-plot. Want curve with non-negative slope. Objects 9 & 11 meet requirement.

h) None of them. Each curve has at least one point with negative velocity (negative value on �-plot).

i) Monotonically changing speed means speed implies one of the following two conditions must be met

• Either the object never slows down.

• Or the object never speeds up.

Object 9 has constant speed and thus meets either requirement.

Object 9 is the only object that meets either requirement.

WATCH OUT! Object 11 slows from speed 15.0�� to 0�� (at 2.50s) then speeds back up to speed 15.0�� .

38

2.29PÔPS Watch out! I switched back to position versus time instead of velocity versus time.

a) Take the absolute value of the initial slope of curve on �-plot. Ranking: �F�* < �Fv* < �F�* < �F�*. b) Want final slope of curve on �-plot. Ranking least negative (or most +) to most negative (least +).��F�) > ��Fv) > ��F�) > ��F�)

c) Linear �-plots have �+ = 0; parabolic plots have �+ = ¬��# ¥ 0.

Objects 14, 15 and 16 meet the requirement.

d) Slope of � should be constant OR becoming more negative (concave downward plots).

Objects 13, 14 and 15. Note: object 13 has �+ = 0 and technically does meet the stated requirement.

e) Want �-plot with non-positive slope. Objects 13 & 15 meet requirement.

f) Object 13.

g) Reverse direction implies velocity changes sign. Velocity is slope of �-plot. Looking for objects with

slope changing sign. Objects 14 & 16 meet the requirement.

h) Don’t over think it; on an �-plot, position is simply the value of the curve (don’t find slope or area).

Displacement is final position minus initial position.

Object 13 displacement is Δ��Fv = 1−15.0m)�̂ − 1+10.0m)�̂ = −25.0m�.̂ Object 14 displacement is Δ��F� = 1−20.0m)�̂ − 15.00m)�̂ = −25.0m�.̂ Object 15 displacement is Δ��F� = 1−17.0m)�̂ − 115.0m)�̂ = −32.0m�.̂ Object 16 displacement is Δ��F� = 15.0m)�̂ − 120.0m)�̂ = −15.0m�.̂ The displacement ranking (most positive to most negative) is Δ��F� > Δ��Fv = Δ��F� > Δ��F�.

i) Distance traveled requires extra thought for any objects which reverse direction (objects 14 & 16).

Split object 14 displacement into chunks: Δ��F�)H|@→F.Ö� = +8.5m� ̂while Δ��F�)H|F.Ö→�� = −33.5m�.̂ Object 14 travels distance ‖Δ��F�‖ = 42m.

Split object 16 displacement into chunks: Δ��F�)H|@→v� = −24.0m� ̂while Δ��F�)H|v→�� = +9.0m�.̂ Object 16 travels distance ‖Δ��F�‖ = 33m.

For objects with monotonically changing position, distance traveled is magnitude of displacement.

The distance ranking is therefore ‖Δ��F�‖ > ‖Δ��F�‖ > ‖Δ��F�‖ > ‖Δ��Fv‖.

I would also accept ‖Δ��F�‖ > ‖Δ��F�‖ D ‖Δ��F�‖ > ‖Δ��Fv‖ for full credit.

Any reading of the graph has some error.

These particular graphs are hard to read beyond the 1’s digit (for values between gridlines).

Within an error tolerance of ±1m, I agree ‖Δ��F�‖ D ‖Δ��F�‖.

39

2.30

a) The plot is shown below. The colors correspond to the various areas used in the displacement calculation (part d).

Be careful, at 2.00 seconds we know velocity is −2.00�� while the speed is +2.00�� .

At the end of this stage the object has three times the speed but moves in the opposite direction.

Opposite direction implies velocity has changed sign to positive.

This, in turn, implies acceleration must be positive.

This tells us the slope of the �-plot is positive after 2.00 seconds.

We keep this positive slope until we reach three times the initial speed (3 × 2.0�� = 6.0�� ).

This occurs when = 5.00s.

b) The magnitude of the acceleration is greatest when the magnitude of the slope is greatest. This occurs between

5.00 & 6.00 seconds. Generally, if a problem statement gives you decimal numbers with units, your answers

should have decimal numbers with units. The default is three sig figs unless otherwise specified.

c) The velocity is zero on a �-plot whenenver the line crosses the time aixs (horizontal axis).

At times 2.75 s & 6.00 s.

d) Total displacement is given by total area under the cure. Areas below the horizontal time axis are negative.

ÌGHG}~ = 5−2.00ms 6 12.00s) + 12 5−2.00ms 6 10.75s) + 12 5+6.00ms 6 12.25s) + 12 5+6.00ms 6 11.00s) ÌGHG}~ = 5.00m

The problem asked for displacement and so we are done. For clarity, let us emphasize the vector nature of this

answer by adding in the appropriate unit vector: Displacement = Δ�ÞÞÞÞ� = 5.00m� ̂e) Final position is found Displacement = ­��A#�inposition Δ�ÞÞÞÞ� = �)ÞÞÞ� − �ßÞÞÞ� �)ÞÞÞ� = �ßÞÞÞ� + Δ�ÞÞÞÞ�

One sees the final position vector equals the displacement vector whenever the initial position vector is zero.

Said another way, whenever and object has initial position at the origin of your coordinates, the displacement

vector equals the final position vector.

Problem continues on next page…

-6

-4

-2

0

2

4

6

0 1 2 3 4 5 6

v (m/s)

t (s)

40

f) The object starts at the origin and begins its motion with negative displacement. We also know the total

displacement is positive from part c. Clearly, at some point the object must transition from negative to positive

net displacement at some point in time. When the object has zero net displacement it is necessarily back at its

initial position (in this case the origin). By looking at the graph, one can see the total area under the curve (total

displacement) should be zero sometime between 4 and 5 seconds.

The area under the curve for the first 2.75 seconds is Ì@Y..Ö��àá = −4.75m

The area between 2.75 seconds and the time of interest is

Ì..Ö�YG = 12 1µ�¬�)1ℎ��#ℎ) Ì..Ö�YG = 12 1Δ)1¬�Õ� ∙ Δ) Here Δ = − 2.75s. You could plug this in now but it is cleaner to leave it as Δ for now.

Ì..Ö�YG = 12 ¬�Õ� ∙ 1Δ).

Ì..Ö�YG = 12 Z��¬��BA[ ∙ 1Δ).

Ì..Ö�YG = 12k8.00ms3.00s l ∙ 1Δ).

Ì..Ö�YG = 12 52.667 ms.6 ∙ 1Δ).

This area must be +6.00m if the object is to return to the origin. 12 52.667 ms.6 ∙ 1Δ). = +5.00m Δ = 1.936s Don’t forget…this is not the time when the object reaches the origin.

It is the time elapsed after = 2.75s when the object reaches the origin. Therefore the object is at the origin at � = �. S�SU

g) Whenever you have a multi-stage equation, the trick is trying to accommodate the nonzero initial time. In

general, for an object starting at time = 0 we know �) = �* + � If the initial time is non-zero, simply shift → − *. �) = �* + �1 − *2

Stage 1 (* = 0.00s2 Stage 2 (* = 2.75s2 Stage 2 (* = 4.00s2 �) = �* + �1 − *2 �) = −2.00ms

�) = �* + �1 − *2 �) = −2.00ms + 2.667 ms. 1 − 2.75s2 �) = �* + �1 − *2 �) = 6.00ms − 6.00 ms. 1 − 5.00s2

41

2.31 The problem statement gives us ��* = +2m� ̂and ��* = −2�� �.̂

x (m)

t (s)

5

-5

-10

-15

-20

-25

-30

0

a (m/s2)

2

-2

0 t (s)

v (m/s)

2

-2

-4

-6

-8

-10

-12

0 t (s)

Abigail x (m)

t (s)

12

-2

6

4

2

8

10

0

a (m/s2)

2

-2

0 t (s)

v (m/s)

Bernie x (m)

t (s)

a (m/s2)

4

-4

0 t (s)

v (m/s)

Cuba Gooding Jr.

t (s)

12

-2

6

4

2

8

10

0

24

-4

12

8

4

16

20

0

t (s)

12

-2

6

4

2

8

10

0

42

2.32

You should be able to read these graphs to get the answers to all parts. If you got the general shape of the graphs

correct that is really what I was looking for. Hopefully your numbers are within about 10% of the chart’s numbers.

0

5

10

15

20

25

30

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0

x (m)

t (s)

0

1

2

3

4

5

6

7

8

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0

v (m/s)

t (s)

0.0

0.5

1.0

1.5

2.0

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0

a (m/s2)

t (s)

43

2.33

a) Moving right implies positive velocity (positive slope on xt plot). Constant speed for a 1D motion implies

acceleration is zero (no concavity). This occurs from about 0 to 3.00 seconds. Note: units are expected for

full credit. Saying 0 to 3 is not good enough. Use three sig figs unless otherwise specified.

b) Moving right implies positive velocity (positive slope on xt). Increasing speed implies acceleration is same

sign as velocity…in this case positive. We are looking for a region with positive slopes and upwards

concavity. This occurs from about 12.5 to 15.0 seconds.

c) Moving right implies positive velocity (positive slope on xt). Decreasing speed implies acceleration is

opposite sign as velocity…in this case negative. In this case we want positive slope but downwards

concavity. This occurs from about 15.0 to 17.5 seconds.

d) Moving left implies negative velocity (negative slope on xt). Constant speed for a 1D motion implies

acceleration is zero (no concavity). This occurs from about 4.00-8.00 sec.

e) Moving left implies negative velocity (negative slope on xt). Increasing speed implies acceleration is same

sign as velocity…in this case negative. We are looking for times with negative slopes and downwards

concavity. This occurs between approximately 17.5 and 20.0 sec.

f) Moving left implies negative velocity (negative slope on xt). Decreasing speed implies acceleration is

opposite sign as velocity…in this case positive. We are looking for times with negative slopes and

upwards concavity. This occurs between approximately 10.0 and 12.5 sec.

g) Not moving implies no velocity (zero slope on xt). This occurs at the times 3.00-4.00 sec, 8.00-10.0 sec, at

12.5 sec and at 17.5 sec.

h) The slope at any point on this curve gives the velocity.

i) The area has units of m ∙ s. This is not a useful quantity for us.

j) The acceleration is positive whenever the xt plot is concave up (between 10.0 and 15.0 sec). The

acceleration is negative whenever the xt plot is concave down (between 15.0 and 20.0 sec). If we ignore

the sharp transitions at 3.00, 8.00, and 10.0 sec, there is no acceleration during the rest of the plot.

k) The velocities desired are given in the table below.

• To get the velocity, use rise over run to get the slope. I recommend drawing the tangent line at

each point and making a little triangle.

• When getting the slope, make the initial point just before the time of interest. Make the final time

just after the time of interest. Using too large of a time interval usually gives bad results.

• Exception to previous comment: if plot is linear, a large time interval works fine.

• On a test I expect your answers to be within perhaps 10% or so of mine.

• I included units in the column headings. Remember to include units on numerical answers for test

questions.

• Remember that velocity has an implied � ̂on it. This means positive answers for velocity imply

motion to the right (forwards) while negative answers for velocity imply motion to the left

(backwards).

Time (sec) Velocity (m/sec)

1.5 2.00

3.5 0

5.0 -0.500

8.5 0

12.5 0

15.0 1.25

17.5 0

20.0 -1.25

l) I haven’t written this up yet part yet. Go on to the next problem.

44

2.34

a) Moving right implies velocity is a positive number (above the time axis). Constant speed implies no

acceleration (no slope on vt plot). This occurs between 0 and 3.00 sec.

b) Moving right implies velocity is a positive number (above the time axis). Increasing speed implies

acceleration is same sign as velocity…in this case positive. We are looking for a region with positive

numbers and positive slopes. This occurs from about 12.5 to 15.0 seconds.

c) Moving right implies velocity is a positive number (above the time axis). Decreasing speed implies

acceleration is opposite sign as velocity…in this case negative. We are looking for a region with positive

numbers and negative slopes. This occurs from about 15.0 to 17.5 seconds.

d) Moving left implies velocity is a negative number (above the time axis). Constant speed implies no

acceleration (no slope on vt plot). This occurs between 4.00 and 8.00 sec.

e) Moving left implies velocity is a negative number (above the time axis). Increasing speed implies

acceleration is same sign as velocity…in this case negative. We are looking for a region with negative

numbers and negative slopes. This occurs from about 17.5 to 20.0 seconds.

f) Moving left implies velocity is a negative number (above the time axis). Decreasing speed implies

acceleration is opposite sign as velocity…in this case positive. We are looking for a region with negative

numbers and positive slopes. This occurs from about 10.0 to 12.5 seconds.

g) Not moving occurs whenever velocity is zero (whenever the line hits the horizontal time axis). This occurs

This occurs at the times 3.00-4.00 sec, 8.00-10.0 sec, at 12.5 sec and at 17.5 sec.

h) The slope has units of |*��|�� = âã � = ��( which are the units of acceleration.

i) The area has units of �� ∙ s = m which are the units of displacement, position, and distance. You are

expected to know the area gives displacement (CHANGE in position).

j) The problem statement tells us “the particle is initially located 3.00 m to the left of the origin”.

This means �* = −3.00m.

We also know the area under the vt curve gives Δ� = �) − �* Rearrange this to make it easy to crank out the final positions. �) = �* + Δ�

The area under the curve for 0-3.00 sec is 6.00 m. Adding this to the initial position of -3.00 m puts the

position at 3.00 s at 3.00 m.

The area under the curve for 3.00-4.00 sec is 0 m. We are therefore still at position 3.00 m.

The area under the curve for 4.00-8.00 sec is -2.00 m. After 8.00 seconds we are thus at 1.00 m.

The area under the curve for 8.00-10.0 sec is 0 m. After 10.0 seconds we are still at 1.00 m.

The area under the curve for 10.0-12.5 sec is -2.00 m. Hopefully your number within 10% of mine. This

means, after 12.5 seconds we are located at -1.00 m.

The area under the curve for 12.5-15.0 sec is 2.00 m. Hopefully your number within 10% of mine. After

15.0 seconds we are once again back at 1.00 m.

The area under the curve for 15.0-17.5 sec is 2.00 m. Hopefully your number within 10% of mine. After

17.5 seconds we are back over to 3.00 m.

The area under the curve for 17.5-20.0 sec is -2.00 m. Hopefully your number within 10% of mine. After

20.0 seconds we arrive once more at 1.00 m.

Remember that position has an implied � ̂on it. This means positive answers for position imply the particle is located

to the right of the origin while negative answers imply the particle is located to the left of the origin.

Problem continues on next page…

45

k) The accelerations desired are given in the table below.

• To get the acceleration, use rise over run to get the slope. I recommend drawing the tangent line at

each point and making a little triangle.

• When getting the slope, make the initial point just before the time of interest. Make the final time

just after the time of interest. Using too large of a time interval usually gives bad results.

• Exception to previous comment: if plot is linear, a large time interval works fine.

• On a test I expect your answers to be within perhaps 10% or so of mine.

• I included units in the column headings. Remember to include units on numerical answers for test

questions.

• Remember that acceleration has an implied � ̂on it. This means positive answers for acceleration

imply motion to the right (forwards) while negative answers for velocity imply motion to the left

(backwards).

Time (sec) Acceleration (m/sec2)

1.5 0

3.5 0

6.0 0

9.0 0

12.5 0.800

15.0 0

17.5 -0.800

20.0 0

l) If you haven’t already realized it, this is the velocity plot for the previous problem! I will show all plots

on the next page so you can see all of them stacked on top of each other. In the acceleration graph I

included some sharp spikes at each time the velocity slope shifts sharply.

46

-3.00

-2.00

-1.00

0.00

1.00

2.00

3.00

0 5 10 15 20

x (m)

t (s)

-1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

2.00

0 5 10 15 20

v (m/s)

t (s)

-2.00

-1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

2.00

0 5 10 15 20

a (m/s2)

t (s)

47

2.35

a) The question says how far during braking for each car. I will use subscript L for the lead car and subscript

T for the trailing car. We know that the lead car comes to rest from speed � with acceleration magnitude �.

In equation form that looks like �n) = �n* − � 0 = � − � = ��

The minus sign comes from moving right and slowing with magnitude �.

This is the time the lead car is braking. Plug this into a distance equation gives

Δ�n = �n* − 12 �.

Δ�n = � 5��6 − 12 � 5��6.

Δ�n = �.2�

Similar work for the trailing car gives �°) = �°* − 14�) 0 = � − 4� = �4�

The minus sign comes from moving right and slowing with magnitude �.

This is the time the lead car is braking. Plug this into a distance equation gives

Δ�ä = �ä* − 12 14�).

Δ�ä = � 5 �4�6 − 2� 5 �4�6.

Δ�ä = �.8�

Lead car brakes for distance «(.} while the trailing car brakes in distance

«(Û}.

WATCHOUT: this is not the total distance driven by the trailing car as we have yet to account for reaction

time.

b) When it comes to rest, the rear bumper of the lead car is distance C + «(.} from the origin. When it comes to

rest, the front bumper of car 2 is distance �| + «(Û} from the origin. Assuming the cars just barely touch,

these two distances must equal. Combining gives | = t« + v«Û}.

c) = «}

d) = «�}

SOLUTION CONTINUES NEXT PAGE…

48

e) See below.

f) See below.

If you have trouble making these plots you can see my data table. I chose the parameters shown at the top. Notice I

made an extra column for on the left side by dividing by the total time 8 s. This gives in terms of a fraction of

the total time ( in units of «}). Notice row 12 is a repeat of the special time |. In cell D12 car 2 starts to decelerate.

The equation is “=$D$11+$D$2*(B12-6)-0.5*$C$2*(B12-6)^2”. Notice the time in these equations is B12-6. This is

like starting the clock over at time | = 6s.

49

2.36

a) Look at that S car go. The solid line indicate the initial conditions while

the dotted lines indicate the conditions at max separation. Max separation

occurs when the two reach the same speed. After that the cop is travelling

faster than the speeder and is closing the gap.

Note: even though the time is unknown, I choose to name it (without a subscript) since it is the same for both cars.

Speeder Cop

∆�F � �*+FF $ 12�+FF. → ∆�F � � At this point I figured there is not much going on here…

I went over to the cop side…

…after figuring out the time from the cop side, I plugged

that in to get

∆�F � �.�

�)+.. � �*+.. $ 2�+.1∆�.2 → �. � 2�∆�

∆�. � �.2�

�)+. � �*+. $ �+.. → � � � � ��

The separation distance is thus

C � ∆�F 0 ∆�. � �.� 0 �.2� � �.2�

Students often mistakenly write down ∆�. � «(.} and think they correctly solved the problem. This is mere

coincidence. Make sure you know what you are doing by trying 2.29 or 2.30 after this.

b) When the cop catches up to the speeder the situation looks like this:

Note: again, time is unknown & I choose to name it ′. BE CAREFUL! Time ’ is not the same as in part a!!!

∆�F � �*+FF $ 12�+FF. → ∆�F � �′ ∆�. � �*+.. $ 12�+... → ∆�. � 12�′.

This time we know the two finish at the same position so we find ∆�F � ∆�.

�′ � 12 �′.

′ � 2��

Using this time gives ∆�F � ∆�. � .«(} and �)+. � 2�. Notice the time to catch the speeder is exactly

twice the time to max separation.

c) If � is larger the cop catches up in less time & over less distance but the final speed is unchanged.

If � is larger the cop catches up in more time & over more distance while final speed increases.

Subtle difference…did you catch it?

Speeder (1) Cop (2) ∆�F ? ∆�. ? �*+F � �*+. 0 �)+F � �)+. � �+F 0 �+. � F ?� . ?�

Speeder (1) Cop (2) ∆�F ? ∆�. ? �*+F � �*+. 0 �)+F � �)+. ? �+F 0 �+. � F ?� ′ . ?� ′

�

�

ç

è

�

� ç

è

�

50

d) The xt plot is shown at right. The

intersection indicates they are at the

same position (the cop has caught up

to the speeder). Notice max separation

occurs at 4 s, half the total time for the

cop to catch the speeder.

e) The vt plot is shown at right. The

intersection indicates when the two

have the same velocity. It is at this

time the cop and speeder have max

separation. After this time the cop

closes the gap to the speeder. Notice

at 8 s the cop has twice the speed of

the speeder but the areas under the

curve (displacements) are identical.

0

80

160

0 4 8

x (m)

t (s)

Speeder

Cop

0

20

40

0 4 8

v (m/s)

t (s)

Speeder

Cop

51

2.37

a) The max separation is no different than the previous problem.

b) In this problem we need to split it into three pieces: one stage for the speeder and two stages for the cop.

Speeder (1) Cop Accel (2) Cop Const § (3) ∆�F ? ∆�. ? ∆�v ? �*+F � �*+. 0 �*+v 1.25� = 54 �

�)+F � �)+. 1.25� = 54 � �)+v 1.25� = 54 � �+F 0 �+. � �+v 0 F ? . ? v ? The � equation for the accelerating cop stage gives . � �«�}.

The �. equation for the accelerating cop gives 5�� �6. � 2�∆�. which gives ∆�. � .�«(v.} .

We also know the following equations relating the givens & unknowns:

The cars travel the same total time when the meet. F � . $ v

The cars have the same displacement when they meet. ∆�F � ∆�. $ ∆�v

The first equation tells us v � F 0 �«�}.

The second gives ∆�F � ∆�. $ ∆�v

�F � 25�.32� $ 54 � ZF − 5�4�[

Solving for F gives F = .�«Û} = 3. 125 «}.

Strictly speaking, the 25% figure probably has 2 sig figs. I wrote 1.25� = �� � for convenience. Please do

not assume the fraction has infinite sig figs.

This is a good time to re-read the question…CRAP…we haven’t solved for wtf yet!

Total distance is easy to find using ∆�F � �F � 3. 125 «(} D 3.2 «(} .

The units check. Also, this distance is 60% longer than problem where the cop accelerated the entire time

while reaching the speeder. Both checks indicate the answer is probably good.

Parts c) and d) on next page…

52

c) For the xt plot I thinned out the lines in an effort to see the intersection more clearly. Notice the

displacement when they meet is larger than 2.36 & 2.38. Seems reasonable if the cop limits the max speed.

d) The vt plot is straightforward since we know all the speeds from the problem statement.

0

80

160

240

0 2 4 6 8 10 12

x (m)

t (s)

Speeder

Cop

0

20

40

0 2 4 6 8 10 12

v (m/s)

t (s)

Speeder

Cop

53

2.38

a) We know they will have same velocity at max separation giving �1 0 ∆2 � � or � «} $ ∆. Speeder travels ∆�F � � � «(} $ �∆. Cop travels ∆�. � F.�1 0 ∆2. � «(.}

Max separation distance is C � «(.} $ �∆. b) Set displacements equal. Find

F. �1 0 ∆2. � �. Rearrange to . 0 25∆ $ «}6 $ 1∆2. � 0.

Solve quadratic to find � .5∆Gw¡̧6±m�5∆Gw¡̧6(Y�1∆G2(. � ∆ $ «} ®1 $ m1 $ .}∆G« ¯

Check the units, good. When ∆ � 0 we get back � .«} just like problem 2.28.

Once is known we could also figure out the distance traveled and final speed of cop.

c) Plot of xt is shown. Checking the quadratic with the provided numbers gives

� 2 $ 205 k1 $ N1 $ 215212220 l � 11.7s Notice the �-intersection occurs just before 12 s on the xt plot!

Notice the max separation distance (the distance between the curves) is greater compared to 2.28.

d) Plot of vt is shown. Notice the final speed is slightly greater than 2� as the cop accelerated for 9.7

seconds instead of only 8 seconds! The �-intersection still occurs 4 seconds after the cop starts

accelerating.

0

80

160

240

0 2 4 6 8 10 12

x (m)

t (s)

Speeder

Cop

0

20

40

60

0 2 4 6 8 10 12

v (m/s)

t (s)

Speeder

Cop

54

2.39

a) Unless otherwise specified, we typically assume SI units. This implies 3�4 = �� .

If units are included the equation gets long and messy so we often do not write them.

That said, unless otherwise specified the implied units are as shown below. �� = é02.00ms $ 58.00ms.6 $ 503.00msv6 .ê �̂ b) Notice velocity has a . term. Upon taking the derivative of ��:

�� = C��C = é8.00ms. 0 6.00msv ê � ̂we see acceleration depends on time. Therefore acceleration is not constant. Therefore it is NOT

appropriate to use the constant acceleration equations of motion.

c) Plug in = 0 to the velocity equation.

The initial VELOCITY is ��Gë@ � 02.00�� �.̂ The initial speed is �Gë@ = ‖��Gë@‖ = 2.00�� .

Speed is the magnitude of velocity.

It has the same units but we do not include the 0 sign or �.̂ d) Acceleration is written in part b. Notice ��@ � 8.00 ��( � ̂while the magnitude is �@ � 8.00 ��(. e) ì� � t}Þ�tG � 06.00 ��í �.̂ Notice jerk is constant.

The initial jerk is ì�@ � 06.00 ��í �.̂ The magnitude is ì@ � 6.00 ��í. f) Acceleration is written in part b. To determine position as a function of time we should integrate.

THINK: because velocity is a function of time only (and not a function of position) there is no need to

use separation of variables. Leaving off units (and �’̂s) to reduce clutter one finds

�) 0 �* � î 102.00 $ 8.00 0 3.00.2CGïG�

�) 0 �* � 102.00 $ 4.00. 0 1.00v2G�Gï

Now assume * � 0 and ) � . Also, from the problem statement the initial position is 02.00m. �) 0 102.002 � 02.00 $ 4.00. 0 1.00v �) � 02.00 0 2.00 $ 4.00. 0 1.00v

Now rename �) as ��, the position as a function of time. Add in the units if desired and an � ̂for clarity. �� � ð02.00 0 2.00 $ 4.00. 0 1.00vñ� ̂g) Moving to the right when �� > 0. Do the quadratic formula on 02.00 $ 8.00 0 3.00. � 0

I found � 0.2792s and = 2.387s. These are the times the velocity plot crosses the horizontal

time axis. We know at time zero the velocity is negative. Therefore it must be positive between these

two times. If you have trouble seeing this, you would simply make a plot of velocity versus time in a

spreadsheet program.

h) Plug in = 3.00s into the above equations to find

• the position �� = 1.00m� ̂• the velocity �� � 05.00�� � ̂ • the SPEED is � � ‖��‖ � 5.00��

• the acceleration �� � 010.00 ��( � ̂

Problem continues on next page…

55

i) Ugh. We know the object is reversing directions at � 0.2792s and = 2.387s from part f.

Determine the position at each of these times so we can record each displacement and take absolute

values. I will ignore the �’̂s for clarity. Notice units are included in the column headings.

1s) �1�) 0 02.00 0.2792 02.268 2.387 2.417 3.00 1.00

Total distance traveled after T. ÑÑU is S. TÜÑ�. Total displacement after T. ÑÑU is T. ÑÑ�ò̂. j) See plots below

-3.00

-2.00

-1.00

0.00

1.00

2.00

3.00

0.0 0.5 1.0 1.5 2.0 2.5 3.0

x (m)

t (s)

-5.00

-4.00

-3.00

-2.00

-1.00

0.00

1.00

2.00

3.00

4.00

0.0 0.5 1.0 1.5 2.0 2.5 3.0

v (m/s)

t (s)

-10.00

-8.00

-6.00

-4.00

-2.00

0.00

2.00

4.00

6.00

8.00

0.0 0.5 1.0 1.5 2.0 2.5 3.0

a (m/s2)

t (s)

Time interval Displacement Δ� Distance |Δ�| 0to0.2792s 00.268m +0.268m 0.2792to2.387s 4.684m 4.684m 2.387to3.00s 01.417m +1.417m

56

2.40 Note: to do calculus with trig functions you typically want to use units of radians.

a) Take derivatives to find � = óÌ sinó and � = ó.Ì cosó. Plug in time = 0 to determine initial values.

We find the initial velocity is zero while the initial acceleration is �*�*G � ó.Ì � 2184 ��(. Notice I converted Ì from cm to m…

b) From experience I know the particle travels distance 2Ì = 10. 0cm before reversing direction. Since

you probably don’t have that fact memorized yet, I figure the easiest thing to do was to make plots and

read the info off the plots. I used a time increment of 1 msec (or 0.001 sec). See plots below.

c) Again, from experience, I know the particle travels distance Ì = 5.0cm before reaching max speed.

Again, I wouldn’t expect you to have this memorized until after an oscillations course.

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0 0.01 0.02 0.03 0.04 0.05

x (m)

t (s)

-15

-10

-5

0

5

10

15

0 0.01 0.02 0.03 0.04 0.05

v (m/s)

t (s)

-3000

-2000

-1000

0

1000

2000

3000

0 0.01 0.02 0.03 0.04 0.05

a (m/s2)

t (s)

57

2.41 �12 = ô�YõG a) 3ô4 = �� and 3ö4 = F� b) �12 = ô11 0 ö2�YõG and �12 = ôö102 $ ö2�YõG c) When � = 0 the particle is be at a ��*� or ��}+. Setting � � 0 and solving for gives { � {|*G*{}~ � Fõ.

To verify � is a max, do a second derivative test. If the second derivative is negative the plot of x vs t is

concave down at { (acceleration is negative, position is at a max). At {|*G*{}~ = Fõ one finds � = − ÷õ�

indicating � at = { is indeed a max.

This is a good time to re-read the question! We are to find the max position, not just {. Plugging in you

should find ��}+ = ÷õ� D 0.3679 ÷õ .

d) Watch out! We know the object changes direction at time { = Fõ.

Split the problem into two parts.

From 0 → { the displacement is ∆�F � � 5Fõ6 0 �102 � 0.3679 ÷õ.

Here � 5Fõ6 means � at time = Fõ.

From { → ..�@õ the displacement is ∆�. � � 5..�@õ 6 − � 5Fõ6 = −0.1627 ÷õ.

The distance traveled is C � |∆�F| $ |∆�.| � 0.531 ÷õ

58

2.42

a) The units of each term in the equation should be m/s. We assume 16.0 ��( and 02.00 ��ø. b) For small times is a small number but v is extremely small.

For example, if � 0.01s we know v = 0.000001sv.

Terms with high powers of are negligible at small times.

c) The exact opposite is true for large times.

For example, if = 10s we know v = 1000sv.

Terms with small powers of are negligible at large times.

d) We are asked to find acceleration when particle has reached maximum positive displacement.

Maximum positive displacement occurs when an object reverses direction.

When an object reverses direction we know velocity is zero.

Setting velocity to zero gives 16.0 0 2.00v � 0 2.0018.00 0 .2 � 0

This implies EITHER 2.00 � 0 OR 8.00 0 . � 0.

The first eqt’n gives the solution � 0 while the second eqt’n gives � ±√8.00 � ±2.828s. The only term which makes sense is = 2.828s. Think: for small we know the positive term dominates in �1); for small we know � > 0.

For large the negative term dominates; for large we know � < 0.

This implies at some positive time the velocity must change from positive to negative (at � 2.828s). This is when the object reverses direction and changes from moving forwards to moving backwards.

Said another way, this is when the object reaches maximum positive displacement.

Remember, we want to find acceleration (not displacement) at this time.

� = C�C = 16.0 0 6.00.

�� � 2.828s� = 16.0 0 6.00�2.828�.

� = −31. 98 D −32.0ms.

SIDE NOTE: To find displacement, we might want to get the area under the � curve (or integrate).

Being extra cautious, I’d like to point out there is no need for separation of variables as velocity is not a

function of position.

Δ� = î 116.0 0 2.00v2C..Û.Û@

Δ� = ù8.00. 0 0.500�ú@

..Û.Û

� = TQ. �

59

e) Taking the derivative of velocity gives acceleration. Setting the derivative (in this case the acceleration)

equal to zero and solving for gives the time of any potential min’s or max’s.

� = C�C = 16.0 0 6.00.

Set equal to zero and solve for time. I found � ±mF�.@�.@@ = ±1.633s. I will ignore the negative time as,

for most problems, we restrict ourselves to positive times only.

To determine if it is a min or a max you could do the second derivative test. The second derivative is

ì = C�C = 012.0 Notice the 2nd derivative is negative for all positive times. This implies the � plot is concave down for all

positive times. This implies the critical point of � 1.633s corresponds to a max velocity. Plugging this

time into the velocity equation gives a max velocity of �� = 17.42 �� �.̂

Watch out! The above answer was max velocity…not max speed!

If we restrict ourselves to positive times, the minimum speed is zero at time = 0. We also know, as time

progresses, the velocity becomes ever more negative. Therefore, at large times the object is always

speeding up! Strictly speaking, there is no max speed!

If you did the above process figuring out the max velocity, don’t feel bad. I did it as well. Only after

I re-read the problem did I notice I asked for speed. It was probably good practice anyways.

f) I made plots on the next page using integration shown below.Δ� = î 116.0 0 2.00v2CGï

@ �) 0 �* � 8.00). 0 0.500)�

After integration, switch ) → . Plug in �* = −8.00 from wording in problem statement.

�) = −8.00 + 8.00. − 0.500�

60

-125.0

-100.0

-75.0

-50.0

-25.0

0.0

25.0

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

x (m)

t (s)

-200.0

-150.0

-100.0

-50.0

0.0

50.0

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

v (m/s)

t (s)

-150.0

-125.0

-100.0

-75.0

-50.0

-25.0

0.0

25.0

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

a (m/s2)

t (s)

61

2.43 You should find the equations of motion are given by: �() = 4 − 3

�() = −2 + 4 − 32 .

�() = −2 + 2. − 12 v

To be thorough, you should be able to determine the appropriate units for each of the numbers in the above

equations. From these you can show the graphs should appear as shown below.

Random comments:

• Using a computer to create plot is so much easier, right? Good skill to learn.

• Using a computer is not a miracle pill which can be used to avoid learning. In many math models there are

tricky spots where the function blows up. Other times the function still gives a number but the underlying

model doesn’t apply to reality anymore. You still need to understand what is going on at a basic level to

find these trouble spots.

• GRAVEL QUARRY RULES!

-8

-7

-6

-5

-4

-3

-2

-1

0

0 0.5 1 1.5 2 2.5 3 3.5 4

x (m)

t (s)

-10

-8

-6

-4

-2

0

2

0 0.5 1 1.5 2 2.5 3 3.5 4

v (m/s)

t (s)

-8

-6

-4

-2

0

2

4

0 0.5 1 1.5 2 2.5 3 3.5 4

a (m/s2)

t (s)

62

2.44

a) The horizontal axis represents time and should have units of seconds.

The vertical axis represents acceleration and should have units of m/s2.

b) For a 1D motion problem, constant velocity implies zero acceleration. This occurs at 3.00 sec and 12.0 sec.

Remember, on test day be sure to include the units on numerical answers! Because, in this case, the

velocity is only constant for two instants out of 16.0 total seconds, one could reasonably argue the velocity

is changing essentially the entire time. That said, notice the acceleration is approximately zero for a few

milliseconds both before and after 12.0 sec. One could argue convincingly the velocity is approximately

constant least at 12.0 sec for a few milliseconds.

c) We know the area under the at curve gives CHANGE in velocity. Δ� = �) − �* The question tells us the final velocity. We can find the area. Rearrange the above equation to solve for �*. �* = �) − Δ�

The area under the curve has units of m/s as expected. I found �* = 10.0 �� . The positive answer indicates

the motion is to the right.

d) First I would split the thing into three stages and write the acceleration as a function of time for those

stages. Notice, in this case, � is a function of time only. I will ignore units, sig figs, and � ̂for clarity.

0-3 sec 3-4 sec 4-16 sec

� = 2 � = 2 + �* � = 2 + 10

Notice the velocity

after 3 seconds is 16.

� = −4 � = −4′ + �* ′ � = −4( − 3) + 16 � = −4 + 28

To get the function correct:

• shift → − 3

• �* of this stage is �) from

previous stage

Notice the velocity after 4 seconds

is 12.

� = −4 + 0.5 � = 0.25′′. − 4′′ + �* ′′ � = 0.25( − 4). − 4( − 4) + 12 � = 0.25. − 6 + 32

To get the function correct:

• shift → − 4

• �* of this stage is �) from

previous stage

Notice the velocity after 16 seconds is 0.

More on next page…

-4.0

0.0

4.0

8.0

12.0

16.0

0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00

v (m/s)

t (s)

63

e) To make the �-plot keep grinding away. I would integrate the velocity equations. Notice I don’t need to

separate the variables as velocity is a function of time only. Also, I assumed the initial position was at the

origin. I will integrate the velocity equations using the formulas with primes and then shift the times.

0-3 sec 3-4 sec 4-16 sec

� = 2 � = 2 + 10 � = . + 10 + �* � = . + 10 + 0

Notice the velocity

after 3 seconds is 16.

Notice the position

after 3 seconds is 39.

� = −4 � = −4′ + �* ′ � = −2û. + �*′′ + �*′ � = −2( − 3). + 16( − 3) + 39 � = −2. + 28 − 27

To get the function correct:

• shift → − 3

• �* of this stage is �) from

previous stage

• �* of this stage is �) from

previous stage

The velocity after 4 seconds is 12.

The position after 4 seconds is 53.

� = −4 + 0.5 � = 0.25′′. − 4′′ + �* ′′ � = 0.08333′′v − 2ûû. + 12ûû + �*′′ � = 0.08333v − 3. + 32 − 32.333

Scratch work shown below

To get the function correct:

• shift → − 4

• �* of this stage is �) from

previous stage

• �* of this stage is �) from

previous stage

� = 0.08333( − 4)v − 2( − 4). + 12( − 4) + 53 � = 0.08333(v − 12. + 48 − 64) − 2(. − 8 + 16) + 12( − 3) + 53 � = 0.08333v − . + 4 − 5.333 − 2. + 16 − 32 + 12 − 48 + 53 � = 0.08333v − 3. + 32 − 32.333

WATCH OUT! At first I simply integrated the simplified versions of the velocity equations to find

0-3 sec 3-4 sec 4-16 sec � = 2 + 10 � = . + 10 + �* � = . + 10 + 0

� = −4 + 28 � = −2. + 28 + �* � = −2. + 28 + 39

� = 0.25. − 6 + 32 � = 0.8333v − 3. + 32 + �* � = 0.8333v − 3. + 32 + 53

Notice these second formulas do not match up at the boundary times. They must be incorrect!

Note: another way to do this might be to code it in Python. Starting from time zero, use the known initial

acceleration, velocity, and positon to compute the velocity and position a tiny time interval later. I would use a tiny

time step (say C = 0.01s). Use these new positions and velocities to predict the next position and velocity. Keep

going iteratively until you reach a transition. Then switch the acceleration and repeat until you hit 16 s.

0.0

25.0

50.0

75.0

0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00

x (m)

t (s)

64

2.45

a) = «} ü1 K m1 0 .t}«( ý b) Use ∆� � F.�.

c) For term in radical to be positive show � > √2C�. Said another way, if term under radical is negative the

solution will not occur. This implies man doesn’t catch bus. One quick way to look at it: if numerator of

fraction is bigger than denominator the second term is larger than 1 and the term under radical is negative.

Man will not catch bus if � > «(.t.

Note: for a realistic problem you might consider the case v = 6 m/s, d = 15 m, and a = 1 m/s2.

2.46 For this problem it is easier to sketch by hand than to use Excel. Note: there is obviously a short time spent

accelerating for the participants of the race to change speed from 2 to 4 m/s. I am assuming this time is negligible.

If this is true, we may assume there is zero acceleration for both stages of the race for both participants.

For Carl For Jane

Half the distance is 12 m.

Since � � 0 we may use C�¬�A­� � ��� > ���

The rate for the first half of the

distance is 2 m/s.

The time spent walking is 6 s.

The time spent running is 3 s. GHG}~ � 9s

Since � = 0 we may use C�¬�A­� � ��� > ���

In this case we know each rate is used for the same amount of time giving

24m = 52ms 6 $ 54ms 6 Notice in this equation the first term on the right-hand side is the distance

walked by Jane while the 2nd term is the distance run by Jane. Also, note

that is not the time she takes to finish…her total time is GHG}~ = 2 = 8s

Note: Why Carl and Jane? Check out Be-Bop Tango by Frank Zappa on the album entitled “Live at the Roxy and

Elsewhere”. Link your mind with the mind of George Duke…

…no, no, you’re still too adagio!

0

5

10

15

20

25

0 1 2 3 4 5 6 7 8 9

x (m)

t (s)

Carl

Jane

0

1

2

3

4

0 1 2 3 4 5 6 7 8 9

v (m/s)

t (s)

Carl

Jane

65

2.47 Not solved yet. The idea is to first determine the time for ball one to fall 10.0 m. Also determine the time for

ball one to rise to 5.00 m and return to earth. Next, we know the time for ball 1 to make the first fall, contact the

ground, then do the entire 5.00 m bounce must equal the time of the second ball to fall 10.0 m PLUS the 1.75 second

time delay. Said another way F@.@�)}~~ + {H�G}{G + �.@m³H��{� = 1.75s + F@.@�)}~~ Notice the time for the 10.0 m fall ends up cancelling out! One finds {H�G}{G = �.@m³H��{� − 1.75s

I will assume up is the positive direction and �� = −#.

For a ball leaving the ground and reaching a max height of 5.00 m bounce we know �)�. = �*�. + 2��Δ? 0 � �*�. $ 210#2� 0�*�. � 210#2�

WATCH OUT! Remember 0�*�. � 0��*��.. The minus sign is not canceled by the squaring process…it cancels

due to the negative sign on �� � 0#. �*�. � 2#� �*� � ±J2#�

Here use the positive root as initially ball is moving upwards from the ground. �)� � J2#�

Now determine time of entire flight. Note: in this special case you could find time to max height and double it since

this problem is a level ground problem. That is how I choose to do it. �)� � �*� $ �� �}+ � �)� 0 �*���

�}+ � 0 0 ��*����

�}+ � 0�J2#��0#

�}+ � J2#�# � N2#�#. � N2�#

Now use

�.@m³H��{� = 2�}+ = 2N2�# = N8ℎ# = N815.00m29.8 ms. = 2.020s Now use {H�G}{G = �.@m³H��{� − 1.75s {H�G}{G = 2.020s 0 1.75s {H�G}{G = 0.270s Notice we lose a sig fig. This is not uncommon when you subtract two numbers close to the same size.

66

2.48 Not solved yet.

2.49 I’m going to be lazy with sig figs on this one.

a) The shortest distance is to run in a diagonal line straight from start to finish. This is a distance of about

22.4 km. This is a bad idea as you do the entire race at the slow speed of 15 km/hr and the race takes

about 89.4 minutes (1.49 hrs).

b) Going straight to the road gives you max possible time on the bike but it will not give you the minimum

total time. Even if you angle slightly over (so b is a small number of km) we know you will run for

essentially the same amount of time yet cycle for less time. The time to run straight down is 40 min plus 20

min to bike over giving a total time of 1 hr for this strategy for the race.

c) The givens are � = 10km, C = 20km, �F = 15 þ�¹º , and �. = 60 þ�¹º . The unknowns are µ and ­.

d) The distance for the first leg is VF � √�. $ µ. while the distance for the second leg is V. � C 0 µ. The

time for the first leg is thus F � n�«� � J}(w³(«� while the time for the second leg is . � n(«( � tY³«( . The total

time to run the race is thus � F $ .

� √�. $ µ.�F $ C 0 µ�.

We want to minimize race time. We are free to change parameter µ. Therefore, take the derivative of with respect to µ, set the derivative to zero, and solve for µ. this value of µ will give a min or a max. Here

it seems fairly obvious we will get a min; if you feel unsure, you could always check with the second

derivative test. CCµ � CCµ Ê 1�F 1�. $ µ.2F/. $ 1�. 1C 0 µ2Ë CCµ � 1�F Z12[ 1�. $ µ.2YF.12µ2 0 1�.

0 � µ�F√�. $ µ. 0 1�.

µ�F√�. $ µ. � 1�.

µ � �F√�. $ µ.�.

µ. � �F.�.. 1�. $ µ.2 µ. � �F.�.. �. $ �F.�.. µ.

µ. ®1 0 �F.�..¯ � �F.�.. �.

µ.1�.. 0 �F.2 � �F.�.

µ � N �F.�.. 0 �F. � D 0.2582�

On this last step I plug in the given values of each � to make it simpler to read.

X = tanYF Zµ�[ ≈ tanYF(0.2582) = 14.5°

67

2.50 The plots shown use the equations derived in problem 2.57. I included the vt plot for ease of reference.

b) Use the slope to get the acceleration. To get the initial acceleration I used

��¬��BA ≈ ��) − ��*) − * = 52 ms �̂6 0 50ms �̂610.22s2 0 10s2 = 9.1 ms. � ̂Typically this gives decent result for the = 0.11¬ halfway between * and ).

c) At the end the slope is almost 0 so the acceleration is almost zero.

d) Part d through h. It is convenient to use the area of

a trapezoid formula. An example for 0.75 to 1.50 s

is worked out to show you how I got the

displacements. Ì��� = µ�¬� > 1�����#����#�2 = 11.50s 0 0.75s) × \12 55ms $ 7.5ms 6] D 4.7m

Also, because the initial position is zero, the first displacement gives the position at 0.75 s. After that,

recall the displacement on each interval is the change in position. Show work for full credit on an exam.

I’m expecting your numbers to be within about 10% of my numbers.

j) A common misconception is to think the marble is slows down at the end. In reality, the marble is

speeding up but at a slower and slower rate. The magnitude of acceleration is decreasing. Eventually, at

terminal velocity, the rate of speed increase is so small we assume speed is constant. It never slows down!

0.0

10.0

20.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0

x (m)

t (s)

x vs t

0.0

5.0

10.0

0.00 0.50 1.00 1.50 2.00 2.50 3.00

v (m/s)

t (s)

v vs t

0.0

5.0

10.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0

a (m/s2)

t (s)

a vs t

time interval ∆x (m) t (s) x (m)

0.00 to 0.75 s 1.9 0.00 0.0

0.75 to 1.50 s 4.7 0.75 1.9

1.50 to 2.25 s 6.1 1.50 6.6

2.25 to 3.00 s 6.7 2.25 12.7

Total (0 to 3 s) 19.4 3.00 19.4

68

2.51

a) Neither. Area under xt curve has units of m ∙ s. It is essentially useless for both velocity and acceleration.

b) Displacement is 0.

c) Distance is 16 m.

d) Velocity is slope of xt plot. Velocity is zero when slope is zero at 0, 2, and 4 sec.

e) Velocity constant when slope is constant at times near 1 and 3 sec. Therefore � = 0 at 1 and 3 sec.

f) Moving right when slope is positive between 0 and 2 sec. Moving left (neg. slope) between 2 and 4 sec.

g) Max velocity is at 1 s. Max speed at 1 sec and at 3 sec!

h) Picking times on either side of 1 sec I found ��¬��BA D ��) 0 ��*) 0 * � 102m�2̂ 0 106m�2̂11.3s) − 10.7s) = 6.7ms � ̂The speed is therefore D 6.7��

i) Concave up xt curve indicates positive acceleration between 0 and 1 sec and between 3 and 4 sec.

Concave down xt curve indicates negative acceleration between 1 and 3 sec.

j) Speeding up when steepness of xt slope increases (between 0 and 1 sec & between 3 and 4 sec).

Slowing down when steepness of xt slope decreases (between 1 and 2 & between 3 & 4 sec).

Another way: whenever a and v have same sign it is speeding up; opposite signs slowing down.

If positive slope and concave up, the mass is moving right and speeding up.

If negative slope and concave down, the mass is moving left and speeding up.

k) See the plots below using the parameters mentioned in part l.

-8.0

-6.0

-4.0

-2.0

0.0

0.0 1.0 2.0 3.0 4.0

x (t)

t (s)

-8.0

-4.0

0.0

4.0

8.0

0.0 1.0 2.0 3.0 4.0

v (m/s)

t (s)

-10.0

-5.0

0.0

5.0

10.0

0.0 1.0 2.0 3.0 4.0

a (m/s2)

t (s)

69

2.52

a) Putting in the �’̂s for clarity, use the equation ��1) = −ô�.�.̂ Plugging in numbers gives �� = −500 ��(.

The negative sign appears because the sled moves to the right and acceleration from drag is to the left.

Simply divide � by # to determine the number of #’s. The magnitude is ‖��‖ ≈ 50#.

b) See plots.

0

20

40

60

80

100

0 1 2 3 4 5 6 7 8 9 10

x (m)

t (s)

0

20

40

60

80

100

0 1 2 3 4 5 6 7 8 9 10

v (m/s)

t (s)

-50

-40

-30

-20

-10

0

0 1 2 3 4 5 6 7 8 9 10

a (m/s2)

t (s)

70

2.53

a

ì = C�C

C� = ìC

î C���

= î ìC��

3�4�� = 3ì4�� �) − �* = ì) − ì* Set * = 0, ) = , and �* = �@. Solve for �) . Rename �) = �() giving �() = �@ + ì

b

� = C�C

�@ + ì = C�C

C� = (�@ + ì)C

î C���

= î (�0 + ì)C��

3�4�� = \Z�0 + 12 ì2[]�

�

�) − �* = Z�@) + 12 ì).[ − Z�@* + 12 ì*.[

Set * = 0, ) = , and �* = �@. Solve for �) . Rename �) = �() giving

�() = �@ + �@ + 12 ì.

c

� = C�C

�@ + �@ + 12 ì. = C�C

C� = Z�@ + �@ + 12 ì.[ C

î C���

= î Z�0 + �0 + 12 ì2[ C�

�

3�4�� = \Z�0 + 12 �02 + 1

6 ì3[]��

�) − �* = Z�@) + 12 �@). + 16 ì)v[ − Z�@* + 12 �@*. + 16 ì*v[

Set * = 0, ) = , and �* = �@. Solve for �) . Rename �) = �() giving

�() = �@ + �@ + 12 �@. + 16 ìv

d

If ì = 0 one finds

�() = �@ + �@ + 12 �@.

�() = �@ + �@ �() = �@

Namely, when � is constant we get back the constant acceleration kinematics equations.

71

2.54

a

C�C = �@ cos ó

When a is a function of t (not a function of v) there is no need to separate...just integrate

î C�)* = �@ î cos ó)

*

�) − �* = �@ó sin ó − �@ó sin 0 = �@ó sin ó

Notice I put in both limits for t. You might think this is a waste of time. I know this is a good habit. You

will see why in the next part…Sometimes that zero limit doesn’t just drop out!

Also, any time you are integrating a trig function you want to watch for sign changes and constants inside the

argument of the trig function. It is always worth verifying you did the integral correctly by taking the

derivative to ensure you get back the correct starting function. Typically, I just guess what the answer is then

fix any goofs by doing this kind of check.

Since �*= 0 (started from rest in problem statement)

�() = �) = �@ó sin ó

It is worth it to check the units and that it gives the correct result for = 0. Looks good.

b

C�C = �@ó sin ó

î C�)* = �@ó î sin ó C)

*

�) − �* = − �@ó. cos ó + �@ó. cos 0 = �@ó. (1 − cos ó)

Notice plugging in the = 0 limit gives rise to the term }��(…it doesn’t simply drop out!

From the problem statement we found �* = −V (neg cuz left of origin)

�() = �) = �@ó. (1 − cos ó) − V

It is worth it to check the units and that our answer gives the correct result for = 0. Looks good.

c ì = C�C = CC �@ cos ó = −�@ó sin ó

72

2.55

a) 3ô4 = ��( and 3ö4 = F

�

b) �() = ô(1 − ö)�YõG and ì() = ôö(−2 + ö)�YõG

c) Separating the variables leads to

î C�)* = î ô�YõGC)

*

Using integration by parts with B = ô and C� = �YõGC gives

î C�)* = \− ôö �YõG]

*) + ôö î �YõGC)

*

�) − �* = \− ôö �YõG]*) + \− ôö. �YõG]

*)

We remember to wait until after all integration to plug in the limits of ) = and * = 0. Watch out!

Notice the 2nd term will give a non-zero result when the * = 0 limit is plugged in! Also, the problem

statement indicated �* = 0.

�() = �) = − ôö �YõG + ôö. − ôö. �YõG = ôö. − ôö Z + 1ö[ �YõG

d) In this case the acceleration goes to zero at { = Fõ which implies the velocity is a maximum, not position.

The only times velocity is zero is at = 0 and = ∞. Therefore the particle will never reach a maximum

position. It will asymptotically approach a value determined by plugging in = ∞. From a practical

standpoint then, the max position is �( = ∞) = ÷õ(.

Note: if you are concerned about the term − ÷õ �YõG = Y��G

��  check the limit using l’Hôpital’s rule.

73

2.56

a

Ignore the minus sign as it has no effect on units.

3ô4 = 3�43�.4 =ms.m.s.

= ms. ∙ s.m. = 1m

b

C�C = −ô�.

Need to separate...a is a function of v.

î C��.)

* = −ô î C)*

\− 1�]*) = −ô

\1�]*) = ô

1�) − 1�* = ô

1�) = 1�* + ô

�) = 11�* + ô

WATCH OUT! So many times I have seen people incorrectly say §É = §� + P�� Consider this simple math problem 12 = 13 + 16

Does this mean 2 = 3 + 6??? No way.

Now multiply each term on right side, in both numerator and denominator, by �* �) = �*1 + �*ô

Rename �) as �(), Rename �* as �@, rearrange slightly gives

�() = �@ 11 + �@ô

Notice the fraction has no units. Result gives the correct result for = 0. Looks good.

c

�() = �* + 1ô ln(1 + �@ô)

It is worth it to check the units and that it gives the correct result for = 0. Looks good.

In problem statement �* = 0 so we could rewrite as �() = F÷ ln(1 + �@ô).

If you are curious what the xt, vt, and at plots look like, they look similar to those in the solution for problem 10.5.

Note: as you increase ô the speed drops off more rapidly.

74

2.57

a 3µ4 = ��}« � = þ�∙âã(âã

= kg ∙ ��( ∙ �

� = þ�� …use this to check the units on your results

b

C�C = # − µ� �

C�# − µ� � = C

15− �µ 6

C�5# − µ� �6 = C 1

5− �µ 6

This step is just to simplify a substitution in the upcoming integral…anyway you get it done is fine! C�� − �#µ = − µ� C

î C�� − �#µ

)* = − µ� î C)

* = − µ�

Let B = � − �o³ . Therefore CB = C�.

î CBB�ï

��= − µ� î C)

* = − µ�

Notice the limits change to B* = �* − �o³ = − �o

³ since released from rest. Also B) = �) − �o³ .

Also, an great shortcut to learn is t++

)* = ln|�|*) = ln��)� − ln|�*| = ln +ï+� . Putting it all together gives

ln ��) − �#µ− �#µ � = − µ�

Exponentiate both sides (do an “e to the” both sides). �) < �o³ so the absolute value is unnecessary.

�) − �#µ− �#µ = �Y ³�G

Solving for �) and renaming it �() gives

�() = �#µ Z1 − �Y ³�G[

Verify when = 0 that � = 0. When = ∞ we find �(∞) = �o³ = ����A�����­�?

c

� = C�C = CC \�#µ Z1 0 �Y³�G[] = CC \�#µ 0 �#µ �Y³�G] = #�Y³�G

Since down is the positive direction this makes sense.

When = 0 the marble has minimal drag with essentially no speed. Seems reasonable � = #.

When = ∞ we find �1∞2 = 0. Drag and weight balance and the marble falls with constant speed.

d

C�C = �#µ 0 �#µ �Y³�G

î C�)* = �#µ î C)

* 0�#µ î �Y³�GC)*

Watch out! When you plug in the �� = Ñ limit on the second integral something survives…

�) = �12 = �* $�#µ 0�.#µ. Z1 0 �Y³�G[

Pretty hairy…take the derivative to make sure you didn’t goof. Check the units. Check if * = 0 gives �*. For large t it looks like constant speed because the exponential drops out. Looks good.

75

2.58 Link to an alternative solution method is provided below:

https://youtu.be/rFUK_V4ygQI

Consider a dropped ball that experiences a viscous force proportional to the speed squared. An

FBD showing the forces and coordinate system are shown at right. Useful tip for later: we know

that at some point (when � = 0) the ball reaches terminal velocity given by

�° = m�#µ

The equation of motion is determined by �# 0 µ�. = �� # 0 ô�. = �

The units of � are therefore

3�4 = 3�43§Q4 = P�� ��U

Separating the variables gives C�C = # 0 µ��.

C� = Z# 0 µ��.[ C C�# 0 µ��. = C µ�C = �µ 1C�2�#µ 0 �.

µ�C = C��°. 0 �.

Note: we expect � < �° for all > 0.

I found

�12 = �°�° $ �*�° 0 �* 0 �Y.³«²� G�° $ �*�° 0 �* $ �Y.³«²� G

When the initial velocity is zero we can simplify this to

�12 = �° tanh Zµ�°� [

Note that � > 0 for all t; this makes sense as I rotated my coordinates such that down was positive.

From here you could use C� = �° tanh 5³«²� 6 C to find �12 in similar fashion. I’d look it up or use a program.

If you are making a video (perhaps using the Tracker program), the first frame of the video might not correspond to = 0. It is sensible to introduce a shift in the time coordinate giving

�12 = �° tanh \µ�°� 1 $ ∆2] where ∆ is the delay time between the release of the ball and the first usable frame of the video.

Comments on next page regarding drag constant.

a

mg

bv2

76

Note: the term b is determined by reading/searching the web about resistive forces. One finds

µ � 12��Ì

where D is a dimensionless number called the drag coefficient, ρ is the density of the fluid the object is passing

through, and A is the cross-sectional area of the object.

For a cotton ball moving through air we expect � � �}*| D 1.2 þ��í. If the cotton ball has a radius of about 3 cm the

cross-sectional area is approximately Ì = 3 > 10Yvm.. A perfectly smooth sphere has a � = 0.5 while rougher

surfaces can have numbers as high as 2. I will therefore assume a number of about 1.5 for a cotton ball as surface

roughness is significant. This gives an estimated value of

µ = 12��Ì = 12 Z1.5 × 1.2 kgmv × 3 × 10Yvm.[ ≈ 3 × 10Yv kgm

Challenge: consider throwing a ball upwards with initial speed less than �°. I believe you get tan instead of tanh in the solution. I bet you can web search for a result as this one is pretty common (try “ballwairresist”). Feeling

frisky? Only a few more cases to consider. Thrown down with initial speed greater than �° and thrown upwards

with initial speed greater than �°. Have fun.