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Chapter 4: Motion with a Changing Velocity

•Motion Along a Line

•Graphical Representation of Motion

•Free Fall

•Projectile Motion

•Apparent Weight

•Air Resistance and Terminal Velocity

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§4.1 Motion Along a Line

xavv

tavvv

tatvxxx

xixfx

xixfxx

xixif

2

2

1

22

2

For constant acceleration the kinematic equations are:

2,

,

fxixxav

xav

vvv

tvx

Also:

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Example: In a previous example, a box sliding across a rough surface was found to have an acceleration of -2.94 m/s2. If the initial speed of the box is 10.0 m/s, how long does it take for the box to come to rest?

Know: a= -2.94 m/s2, vix=10.0 m/s, vfx= 0.0 m/s

Want: t.

sec 40.3m/s 942

m/s 0.10

0

2

.-a

vt

tavv

x

ix

xixx

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Example (text problem 4.8): A train of mass 55,200 kg is traveling along a straight, level track at 26.8 m/s. Suddenly the engineer sees a truck stalled on the tracks 184 m ahead. If the maximum possible braking force has magnitude 84.0 kN, can the train be stopped in time?

22

22

m/s 95.12

02

x

va

xavv

ixx

xixx

Know: vfx = 0 m/s, vix=26.8 m/s, x=184 m

Determine ax and compare to the train’s maximum ax.

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Example continued:

The train’s maximum acceleration is:

2max, m/s 52.1

m

F

m

Fa brakingnetx

The maximum acceleration is not sufficient to stop the train before it hits the stalled truck.

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§4.2 Visualizing Motion with Constant Acceleration

Motion diagrams for three carts:

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Graphs of x, vx, ax for each of the three carts

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Example (text problem 4.13): A trolley car in New Orleans starts from rest at the St. Charles Street stop and has a constant acceleration of 1.20 m/s2 for 12.0 seconds.

0

2

4

6

8

10

12

14

16

0 2 4 6 8 10 12 14

t (sec)

v (m

/sec

)

(a) Draw a graph of vx versus t.

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(b) How far has the train traveled at the end of the 12.0 seconds?

The area between the curve and the time axis represents the distance traveled.

m 4.86s 12m/s 4.142

1

tsec 12t2

1

vx

(c) What is the speed of the train at the end of the 12.0 s?

This can be read directly from the graph, vx=14.4 m/s.

Example continued:

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§4.3 Free Fall

A stone is dropped from the edge of a cliff; if air resistance can be ignored, the FBD for the stone is:

x

y

w

Apply Newton’s Second Law

2m/s 9.8

N/kg 8.9

ga

mamgwFy

The stone is in free fall; only the force of gravity acts on the stone.

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Example: You throw a ball into the air with speed 15.0 m/s; how high does the ball rise?

Given: viy=+15.0 m/s; ay=-9.8 m/s2

2

2

1tatvy yiy

x

yviy

ay

tavv yiyfy

To calculate the final height, we need to know the time of flight.

Time of flight from:

12

sec 53.1m/s 9.8-

m/s 0.15

0

2

y

iy

yiyfy

a

vt

tavvThe ball rises until vfy= 0.

m 5.11

s 53.1m/s 8.92

1s 53.1m/s 0.15

2

1

22

2

tatvy yiyThe height:

Example continued:

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Example (text problem 4.22): A penny is dropped from the observation deck of the Empire State Building 369 m above the ground. With what velocity does it strike the ground? Ignore air resistance.

369 m

x

yGiven: viy=0 m/s, ay=-9.8 m/s2, y=-369 m

Unknown: vyf

Use:

yav

ya

yavv

yyf

y

yiyfy

2

2

222

ay

14

m/s 0.85m 369m/s 8.922 2 yav yyf

How long does it take for the penny to strike the ground?

sec 7.82

2

1

2

1 22

y

yyiy

a

yt

tatatvy

(downward)

Example continued:

Given: viy=0 m/s, ay=-9.8 m/s2, y=-369 m Unknown: t

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§4.4 Projectile Motion

What is the motion of a struck baseball? Once it leaves the bat (if air resistance is negligible) only the force of gravity acts on the baseball.

The baseball has ax = 0 and ay = -g, it moves with constant velocity along the x-axis and with nonzero, constant acceleration along the y-axis.

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Example: An object is projected from the origin. The initial velocity components are vix = 7.07 m/s, and viy = 7.07 m/s. Determine the x and y position of the object at 0.2 second intervals for 1.4 seconds. Also plot the results.

tvx

tatatvy

ix

yyiy

22

2

1

2

1

Since the object starts from the origin, y and x will represent the location of the object at time t.

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t (sec) x (meters) y (meters)

0 0 0

0.2 1.41 1.22

0.4 2.83 2.04

0.6 4.24 2.48

0.8 5.66 2.52

1.0 7.07 2.17

1.2 8.48 1.43

1.4 9.89 0.29

Example continued:

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0

2

4

6

8

10

12

0 0.5 1 1.5

t (sec)

x,y

(m

)

This is a plot of the x position (black points) and y position (red points) of the object as a function of time.

Example continued:

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Example continued:

0

0.5

1

1.5

2

2.5

3

0 2 4 6 8 10

x (m)

y (m

)

This is a plot of the y position versus x position for the object (its trajectory).

The object’s path is a parabola.

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Example (text problem 4.36): An arrow is shot into the air with = 60° and vi = 20.0 m/s.

(a) What are vx and vy of the arrow when t=3 sec?

x

y

60°

The components of the initial velocity are:

m/s 3.17sin

m/s 0.10cos

iiy

iix

vv

vv

At t = 3 sec:m/s 1.12

m/s 0.10

tgvtavv

vtavv

iyyiyfy

ixxixfx

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(b) What are the x and y components of the displacement of the arrow during the 3.0 sec interval?

y

x

r

m 80.72

1

2

1

m 0.3002

1

22

2

tgtvtatvyr

tvtatvxr

iyyiyy

ixxixx

Example continued:

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Example: How far does the arrow in the previous example land from where it is released?

The arrow lands when y=0. 02

1 2 tgtvy iy

Solving for t: sec 53.32

g

vt iy

The distance traveled is:

m 35.302

1 2

tv

tatvx

ix

xix

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§4.5 Apparent Weight

Stand on a bathroom scale.

FBD:Apply Newton’s 2nd Law:

y

yy

mamgN

mawNF

w

N

x

y

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The normal force is the force the scale exerts on you. By Newton’s 3rd Law this is also the force (magnitude only) you exert on the scale. A scale will read the normal force.

yagmN is what the scale reads.

When ay = 0, N = mg. The scale reads your true weight.

When ay0, N>mg or N<mg.

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Example: A woman of mass 51 kg is standing in an elevator. If the elevator pushes up on her feet with 408 newtons of force, what is the acceleration of the elevator?

FBD for woman:

y

yy

mamgN

mawNF

Apply Newton’s 2nd Law:

(1)w

N

x

y

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Solving (1) for ay:2m/s 8.1

m

mgNay

The elevator could be (1) traveling upward with decreasing speed, or (2) traveling downward with increasing speed.

Given: N = 408 newtons, m = 51 kg, g = 9.8 m/s2

Unknown: ay

Example continued:

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§4.6 Air Resistance

A stone is dropped from the edge of a cliff; if air resistance cannot be ignored, the FBD for the stone is:

Apply Newton’s Second Law

mawFF dy x

y

w

Fd

Where Fd is the magnitude of the drag force on the stone. This force is directed opposite the object’s velocity

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Assume that 2bvFd

b is a parameter that depends on the size and shape of the object.

Since Fdv2, can the object be in equilibrium?

b

mgvv

mgbv

mawFF

t

dy

when yes,

02

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Example (text problem 4.61): A paratrooper with a fully loaded pack has a mass of 120 kg. The force due to air resistance has a magnitude of Fd = bv2 whereb = 0.14 N s2/m2.

(a) If he/she falls with a speed of 64 m/s, what is the force of air resistance?

N 570m/s 64/ms N 14.0 2222 bvFd

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(b) What is the paratrooper’s acceleration?

Apply Newton’s Second Law and solve for a.

2m/s 1.5

m

mgFa

mawFF

d

dy

x

y

w

FdFBD:

(c) What is the paratrooper’s terminal speed?

m/s 92

0

02

b

mgv

mgbv

mawFF

t

t

dy

Example continued:

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Summary

•The Kinematic Equations

•Graphical Representations of Motion

•Applications of Newton’s Second Law & Kinematics (free fall, projectiles, accelerated motion, air drag)

•Terminal Velocity