1 topics distance, location, speed speed and direction directional quantities acceleration free fall...

1

Topics

• Distance, Location, Speed

• Speed and Direction

• Directional quantities

• Acceleration

• Free Fall

• Graphs of Motion

• Derivatives and Integrals

2

Average Speed

• distance: total path length

• speed: rate of travel (e.g. 50 mph)

• Average Speed: distance/time (e.g. 100m in 3.0s)

[m/s] timetravel

distances

3

Displacement: Change in Position

if xxx SI Unit: meters (m)

4

Velocity (m/s)

0 : : velocityaverage

tt

xvavg

0 : : velocityousinstantane

tt

xv

5

Velocity Examples

• average velocity: 60mph toward Dallas

• instantaneous velocity: 11:47am: Northbound, 83mph

6

Example: Average Velocity

to = 0.0s, xo = 5.0m, vo = +2.0m/s

t = 1.2s, x = 3.08m, v = -5.2m/s

smss

mm

t

xvavg /6.1

0.02.1

00.508.3

Note that velocities always have directional information. Here the “-” sign means –x direction.

7

Scalars & Vectors

• Scalar: size only

• e.g. speed, distance, time

• Vector: magnitude and direction

• e.g. displacement, velocity, acceleration

8

A honeybee travels 2 km round trip before returning. Is the displacement for the trip the same as the distance traveled?

1 2

79%

21%

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

41 42 43 44 45

1. Yes

2. No

9

Acceleration (m/s/s)

0 : :onaccelerati average

tt

vaavg

0 : :onaccelerati ousinstantane

tt

va

10

Example: Car goes from 10m/s to 15m/s in a time of 2.0 seconds. Calculate the average acceleration.

m/s/s 5.20.0s-2.0

10m/s-15

t

vaavg

11

Previous Example:

to = 0.0s, xo = 5.0m, vo = +2.0m/s

t = 1.2s, x = 3.08m, v = -5.2m/s

m/s/s 0.60.0s-1.2

2.0m/s-5.2-

t

vaavg

12

Motion Diagrams

• velocity arrow and position• zero velocity is a “dot”• acceleration & net-force directions: parallel to v• Example: slowing, reversing direction

13

Kinematic Equations of Constant Acceleration

atvv o :velocity

tvvx o )( : velocityaverage 21

221 :ntdisplaceme attvx o

xavv o 2 :squared-v 22

14

Displacement and x vs. t Graph

15

x vs. t Graph

• slope is velocity

16

v vs. t Graph• slope is acceleration

atvv o

17

Human Acceleration

mat 20221

In the 1988 Olympics, Carl Lewis reached the 20m mark in 2.96s. Calculate average acceleration.

20)96.2( 221 a

ssms

ma //56.4

)96.2(

2022

18

Cheetah Acceleration

A cheetah can accelerate from 0 to 20m/s in 2.0s. What is the average acceleration?

ssms

sm

t

vva o //10

0.2

/)020(

19

Ex: V2 EquationApproximate Stopping Accelerations in m/s/s:

Dry Road: ~ 9 (anti-lock) ~ 7 (skidding)

Wet Road: ~ 4 (anti-lock) ~ 2 (skidding)

At 60mph = 27m/s, what is the stopping distance of a skid on a wet road?

feet) 006(about 182

)2(2270

222

22

mx

x

xavv o

20

Free-Fall

• only gravity acts

• air-friction is negligible

• a = 9.8m/s/s downward

21

Calculus of Linear Motion

• derivatives and integrals

• Examples:

• dx/dt = v dv/dt = a

• d/dt(3 + 4t + 5t2) = 4 + 10t

• v = integral of acceleration

22

Velocity

a(t)dtvv o

22545 ttdtt

Example:

23

Summary:• speed: rate of travel• average speed: distance/time.• displacement: change in position• velocity: rate position changes• acceleration: rate velocity changes• kinematic equation set• free fall: constant acceleration.• graphs and slopes• derivatives and integrals of polynomials

24

25

Example: A solid metal ball is projected directly upward with velocity +5.0m/s. How high does it go? How long does it take to return to same height?

mh

gh

gh

yavv o

28.1

6.19/25)2/(25

250

222

22

sgt

gt

t

gttgtt

attvy o

02.18.9/10/10

0)5(

0

)5(50

21

212

21

221

26

Case Study: 100 meter track-race

1. a = const., 0-60 m 2. top speed of 16 m/s at 60 m. 3. a = 0, 60-100 m

velocity vs time

0.002.004.006.008.00

10.0012.0014.0016.0018.00

0.00 2.00 4.00 6.00 8.00 10.00 12.00

t(s)

velo

city

(m/s

)

27

st

t

t

at

attvx

t

o

5.78/60

860

60

060216

21

221

221

2/13.25.7/16

16

016

sma

ta

at

atvv o

a) Acceleration and Time

100m Race

28

st

t

attvx o

5.216/40

01640

221

b) Time and Distance: Last 40meters of race at constant speed of 16m/s.

Race Time = tI + tII = 7.5s + 2.5s = 10.0s

100m Race

29

v = vo + at.16 = 0 + a(7.5)a = 16/7.5 = 2.13 m/s2.

c) We can also use time found in part (a) in velocity equation to get the acceleration of the runner in 1st part of the race.

x = vavgt = {(vo + v)/2}t = {(0 + 16)/2)}(7.5) = (8)(7.5) = 60m.

d) Distance using vavg

30

Position vs time

0.00

20.00

40.00

60.00

80.00

100.00

120.00

0.00 2.00 4.00 6.00 8.00 10.00 12.00

t(s)

po

sit

ion

(m)

31

Example: An object has velocity of +2.0m/s at x = 5.0m and at t = 0.0s. At t = 1.2s it has velocity of -5.2m/s and position x = 3.08m.

Average Acceleration:

ssmss

smsm

t

vaavg //0.6

0.02.1

/0.2/2.5

smsssmsmatvv o /2.5)2.1)(//0.6(/0.2

Using v(t) equation:

Consistent answer:

How long did it take the object to reach v = 0?

sssm

sm

a

vt

atv

o

o

33.0//0.6

/0.200

0

32

33

A train moves along a straight track. The graph shows the position as a function of time for this train. Note that the speed at an instant is the slope of the line at any point on the line. The graph shows that the train:

1 2 3 4

11%

22%

39%

28%

1. speeds up all the time.

2. slows down all the time.

3. speeds up part of the time and slowsdown part of the time.

4. moves at a constant velocity.

time

position

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

41 42 43 44 45

34

A car travels West at 20m/s. It begins to slow. Use the convention that East is +x. The acceleration of the car is considered positive since if it slowed to 19m/s in 1.0s, then

ssms

sm

t

vva o //1

1

/)20(19

Motion Diagram:

v

v(t)

a

+-

Motion Diagram Example

35

Example: A car starts from rest and travels West with uniformly increasing speed. Use the convention that East is +x. Is the acceleration + or -? Is the total force acting on the car + or -? Draw a motion diagram.

Assume it goes from 0 to -10m/s in 10s.

ssms

sm

t

vva o //1

10

/)0(10

Net-force parallel to acceleration, i.e. force is – direction.

motion diagram

Net Force, Acceleration, & Motion Diagrams

36

A car can accelerate at 6m/s/s. The time to go from 40mph to 60mph is:

smmi

m

s

h

h

mi/87.17

1

1609

3600

140 sm

mi

m

s

h

h

mi/81.26

1

1609

3600

160

atvv o

sssm

sm

a

vvt o 49.1

//6

/87.1781.26

Example using Acceleration

37

VehicleAverage Stopping Distance at 55 mph (includes reaction time)

Passenger car 190 ft.

Tractor-trailer (loaded) with cool brakes

256 ft.

Tractor-trailer (loaded) with hot brakes

430 ft.

Tractor-trailer (empty) 249 ft.

Tractor only (bobtail) 243 ft.

38

VehicleStopping Distancefrom 60 mi/hr

Accel.

feet meters ft/s2 m/s2

BMW M3 120 37 32.3 9.8

Dodge Colt GL 167 51 23.2 7.1

39

40

Time to Stop

BMW

st

t

atvv o

75.2

8.9270

st

t

atvv o

80.3

1.7270

Colt

41

y and v graphs for tossed object in “free-fall”

42

Determine how realistic 6m/s/s is for a car by computing the 0 to 60mph time:

Good time, but can be done.

sssm

sm

a

vvt o 46.4

//6

/081.26

Realistic Car?