physics for scientists and engineers - · pdf filespring, 2008 ho jung paik physics for...

Spring, 2008 Ho Jung Paik

Physics for Scientists and Engineers

Chapter 2Kinematics in One Dimension

27-Jan-08 Paik p. 2

KinematicsDescribes motion while ignoring the agents (forces) that caused the motion

For now, will consider motion in one dimension

Along a straight line

Will use the particle modelA particle is a point-like object, has mass but infinitesimal size

27-Jan-08 Paik p. 3

Acceleration and Velocity

27-Jan-08 Paik p. 4

Motion Diagrams

Position graphs ⇒

27-Jan-08 Paik p. 5

Interpreting a Position Graph

27-Jan-08 Paik p. 6

Positive & Negative Slopes

27-Jan-08 Paik p. 7

Uniform & Nonuniform Motion

27-Jan-08 Paik p. 8

Uniform Motion

tvxxtxx

txv

xif

ifx

Δ+=Δ−

=ΔΔ

≡

A motion with

Consider 1-D motion in x direction

constant=vr

27-Jan-08 Paik p. 9

Collision ProblemBob leaves home in Chicago at 9:00 am and travels east at a steady 60 mph. Susan, 400 miles to the east in Pittsburgh, leaves at the same time and travels west at a steady 40 mph. Where will they meet for lunch?

27-Jan-08 Paik p. 10

Position vs Time and Math

1

0101

11

0101

mi 400 )(

)h 0(mi 0 )(

tvttvxx

tvtvttvxx

S

SSS

BB

BBB

+=−+=

=−+=−+=

h 0.4)]mph 40(mph 60[

mi 400mi 400)(

mi 400

1

11

11

11

=−−

=

=−+=

=

t

tvtvtvtv

xx

SB

SB

SB

SB

BB

xxtvx

11

111

mi 240h 0.4mph 60

==×==

They meet at t = t1 and when x1B = x1S.

Using this t1 in Bob’s equation

h 0.4)]mph 40(mph 60[

mi 400mi 400)(

mi 400

1

11

11

11

=−−

=

=−+=

=

t

tvtvtvtv

xx

SB

SB

SB


Instantaneous VelocityThe limit of the average velocity as the time interval becomes infinitesimally short.

The instantaneous velocity indicates what is happening at every point of time.

dtdx

txv

tx =ΔΔ

=→Δ 0

lim

Slope of the green line


Instantaneous Velocity, cont

The instantaneous speed is the magnitude of the instantaneous velocity

The average speed is not always the magnitude of the average velocity!


Position from Velocity

dtdxvx =

dtvxx

dtvdx

dtvdx

t

t x

t

t x

x

x

x

∫

∫∫=−

=

=

1

0

1

0

1

0

01

Since the instantaneous velocity is

the change in position of a moving object is given by


Motion Equation from Calculus

Displacement equals the area under the velocity–time curve

The limit of the sum is a definite integral:


Instantaneous AccelerationInstantaneous acceleration is the limit of the average acceleration as Δt approaches 0:

2

2

0

, Since

lim

dtxda

dtdxv

dtdv

tva

xx

xx

tx

==

=ΔΔ

=→Δ


Instantaneous Acceleration, contThe slope of the velocity vs time graph is the acceleration

The blue line is the average acceleration between ti and tfThe green linerepresents the instantaneous acceleration at tf


Acceleration and Velocity, 1

The car is moving with constant positive velocity (red arrows maintaining the same size)

Acceleration equals zero

Equal time delay snapshots



Velocity and acceleration are in the same direction

Acceleration is positive and uniform (blue arrows maintaining the same length)

Velocity is positive and increasing (red arrows getting longer)




Acceleration and velocity are in opposite directions

Acceleration is negative and uniform (blue arrows maintaining the same length)

Velocity is positive and decreasing (red arrows getting shorter)



Constant AccelerationFor constant acceleration, the average velocity can be expressed as the arithmetic mean of the initial and final velocities

Not true if a is not constant

20 vvvave+

=


Velocity from AccelerationSince the instantaneous acceleration is

the change in velocity is given by

If a is a constant,

∫∫ =t

t

t

tadtvvdv

00

=- 0

dtdva =

)( 00

ttaadtt

t−=∫

)(+= 00 ttavv − Eq. (1)


Displacement from AccelerationFor constant acceleration,

Since and

Therefore,

( )vvvave += 021

)( 00 ttvxx ave −+=

)(+= 00 ttavv −

[ ] )()( 021

000021 ttavttavvvave −+=−++=

202

1000 )()( ttattvxx −+−+= Eq. (2)


With Time EliminatedFrom Eq. (1),

Substituting this into Eq. (2),

Therefore,

avvtt 0

0−

=−

)(2 020

2 xxavv −+=

( )200

2200

20

210

00

22221 vvvvvvva

avva

avvvxx

+−+−=

⎟⎠⎞

⎜⎝⎛ −

+⎟⎠⎞

⎜⎝⎛ −

=−

Eq. (3)


1-D Kinematic EquationsWith constant acceleration,

You may need to use two of the equations to solve one problem

Many times there is more than one way to solve a problem

)(2 )3(

)()( )2(

)( (1)

020

2

202

1000

00

xxavv

ttattvxx

ttavv

−+=

−+−+=

−+=


Displacement–Time CurveSlope of the curve is the velocity

Curved lineindicates the velocity is changing

Therefore, there is an acceleration


Velocity–Time CurveSlope is the acceleration

Straight lineindicates the velocity is changing at a constant rate

Therefore, a constant acceleration


Slope is the rate of change in acceleration

Zero slopeindicates a constant acceleration

Acceleration–Time Curve


Example of a =0A car is moving at a constant velocity of 60 mph. The driver suddenly sees an animal crossing the road ahead. If the driver’s reaction time is 0.20 s, show how far the car will go before the driver pushes on the brake pedal.

202

1000 )()( ttattvxx −+−+=

We knowv0 = 60 mph x (0.447 m/s / 1 mph) = 27 m/s, a = 0 m/s2, t – t0 = treact = 0.20 s

Use the equation:

The displacement of the car before putting on the brakes:x – x0 = 27 m/s x 0.20 s + 0 m = 5.4 m


A particle starts from rest and accelerates as shown in the figure. Determine (a) the particle’s speed at t = 10.0 s and 20.0 s, and (b) the distance traveled in the first 20.0 s.

Example of a ≠ 0

( )( )( ) m 2630.5)00.3(0.50.20200

m 2000.500.50.20100

m 1000.1000.20.1000 (b)

m/s 0.50.5)00.3(0.20 m/s 0.200.500.20 m/s 0.200.1000.20 (a)

2212

3321

3151520

2212

2221

2101015

2212

1121

10010

331520

221015

11010

=×−+×+=Δ+Δ+=

=××+×+=Δ+Δ+=

=××+×+=Δ+Δ+=

=×−+=Δ+==×+=Δ+==×+=Δ+=

tatvxx

tatvxx

tatvxx

tavvtavvtavv


Freely Falling ObjectsA freely falling object is any object moving freely under the influence of gravity alone, i.e., with negligible air drag

Object has a constant acceleration due to gravity

Demonstration 1: Free fall in vacuum

Demonstration 2: Measure a


Acceleration of Free FallAcceleration of an object in free fall is downward, regardless of the initial motion

The magnitude of free fall acceleration is usually taken as g = 9.80 m/s2, however

g decreases with altitudeg varies with latitude9.80 m/s2 is the average at the Earth’s surface


Free Fall Example

At A, initial velocity is upward (+20 m/s) and acceleration is –g (–9.8 m/s2). At B, velocity is 0 and acceleration is –g. At C, velocity has the same magnitude as at A, but is in the opposite direction. The final displacement is –50.0 m. (a) Find the distance from A to B.(b) Find the velocity at C. (c) Find the velocity at yE = –50.0 m.


Free Fall Example, cont

sign) minus thechoose( m/s 1.37 )m/s(1380)m/s(980)m/s(400

)m 0m 0.50)(m/s 80.9(2)m/s 0.20(

)(2 (c)

down) as sign, minus (choose m/s 0.20 )m/s(8.399)m 4.20m 0)(m/s 80.9(20

)(2 (b)

m 4.20m/s 60.18

)m/s(400)m/s 80.9(2

)m/s 020(

)m 0)(m/s 80.9(2)m/s 0.20(0

)(2 (a)

222

222

22

22

22

2

2

2

2

22

22

±==+=

−−−+−=

−+=

±==−−+=

−+=

===

−−+=

−+=

E

E

CECE

C

BCBC

B

B

ABAB

v

vyyavv

v

yyavv

.y

yyyavv


Problem Solving – Conceptualize Think about and understand the situation

Make a quick drawing of the situation

Gather the numerical informationWrite down the “givens”

Focus on the expected resultWhat are you asked to find?

Think about what a reasonable answer should be, e.g., sign of quantities, units


Problem Solving – Categorize Simplify the problem

Can you ignore air resistance? Model objects as particles

Try to identify similar problems you have already solved


Problem Solving – Analyze Select the relevant equation(s) to apply

Solve for the unknown variable

Substitute appropriate numbers

Calculate the resultsInclude units

Round the result to the appropriate number of significant figures


Problem Solving – FinalizeCheck your result

Does it have the correct units?Does it agree with your conceptualized ideas? Does the answer have the correct signs?

Look at limiting situations to be sure the results are reasonable

e.g., the correct limit if a = 0?

Compare the result with those of similar problems


Example 1

A hare and a tortoise compete in a 1.00 km race. The tortoise crawls steadily at its maximum speed of 0.200 m/s toward the finish line. The hare runs at its maximum speed of 8.00 m/s toward the goal for 0.800 km and then stops to tease the tortoise. How close to the goal can the hare let the tortoise approach before resuming the race?


Example 1, cont

The hare sits at 800 m waiting for the tortoise. To go the last 200 m, the hare will takeΔt = 200 m/(8.00 m/s) = 25.0 s

In 25.0 s, the tortoise can goΔx = 0.200 m/s x 25.0 s = 5.00 m

In order for the hare to win, he must restart before the tortoise gets within 5.00 m of the finish line.


Example 2Jules Verne in 1865 suggested sending people to the Moon by firing a space capsule from a 220-m long cannon with a final velocity of 10.97 km/s. What would have been the acceleration experienced by the space travelers during launch? Compare your answer with the free-fall acceleration 9.80 m/s2.

1079.2m/s 10735.2m 440

)m/s(103.120)m 220(2m/s) 0()m/s1097.0(1

)(2

m 220,m 0,m/s 109710 ,m/s 0

42526

223

22

3

ga

a

yyavv

yy.vv

ifif

fifi

×=×=×

=

+=×

−+=

==×==


Example 3

Speedy Sue, driving at 30.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 155 m ahead traveling at 5.00 m/s. Sue applies her brakes but can accelerate only at −2.00 m/s2 because the road is wet. Will there be a collision? If yes, determine how far into the tunnel and at what time the collision occurs.


Example 3, cont

m 212s 4.11m/s 00.5m 155)( :positionat happens wreck The time.collision theis onesmaller The

s 11.4or s 6.132

15540.250.25

01550.25or 00.51550.30

:)()( to solutiona is thereif occurs Collision)m/s 0()m/s 00.5(m 155 )(

:position sVan')m/s 00.2(m/s) 0.30(m 0)(

:position sSue'

2

22

22212

21

00

22212

21

00

=×+=

=×−±

=

=+−+=−

=

++=++=

−++=++=

CV

C

CCCCC

CVCSC

VVVV

SSSS

tx

t

ttttttxtxt

tttatvxtx

tttatvxtx


Example 4A student climbs a 50.0-m cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.00 m/s. (a) How long after release of the first stone do the two stones hit the water? (b) What initial velocity must the second stone have? (c) What is the speed of each stone at the instant the two hit the water?

50. 0 m

0 m0 s 1.00 s

y

t


Example 4, cont

m/s 9.34)s 00.1s 00.3)(m/s 80.9(m/s 3.15

)(v m/s 4.31)s 0s 00.3)(m/s 80.9(m/s 00.2

)( (c)

m/s 2.1500.2

6.190.50

)s 00.1s 00.3)(m/s 80.9()s 00.1s 00.3(m 0m 0.50

)()( :stone second The (b)

) (Choose s 40.3or s 00.3)90.4(2

)0.50)(90.4(400.200.2

)s 0)(m/s 8.9()s 0)(m/s 00.2(m 0.50m 0

)()( :stone First (a)

220202

210101

20

2221

20

2202

12020202

2

2221

2012

10101011

−=−−+−=

−+=−=−−+−=

−+=

−=+−

=

−−+−+=−

−+−+=

+−=−−±−

=

−−+−−+=

−+−+=

ttav

ttavv

v

vttattvyy

t

ttttattvyy


Example 5The Acela is the Porsche of American trains. It can carry 304 passengers at 170 mi/h. A velocity-time graph for the Acela is shown in the figure.(a) Find the peak positive acceleration of the train. (b) Find the train’s displacement between t = 0 and t = 200 s.


Example 5, cont

t (s)100 200 300

–100

100

200

400

Dv

D t–500

0

2m/s 98.0s1

s 3600m 16002.2

(mi/h)/s 2.2s 50)-(100

mi/h )45155(

=××=

=−

=ΔΔ

=tva

(a) Peak acceleration is given by the slope of the steepest tangent to the v -t curve.

From the tangent line shown, we find


(b) Area under the v -t curve equals the displacement.

We approximate the area with a series of triangles and rectangles.

Example 5, cont

t (s)100 200 300

100

200

400

1 24 3

5

00

mi 6.7ss 3600mi 24000smi/h 24000

s 100mi/h )160170(s 50mi/h 100

s 100mi/h 160s 50mi/h 50s 50mi/h 50 5area 4area 3area 2area 1area

21

21

=×=×=

×−×+××=

×+×+×=++++=Δx

physics for scientists and engineers - · pdf filespring, 2008 ho jung paik physics for...

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