1Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 3LectureOutline

2

Chapter 3: Acceleration and Newton’s Second Law of Motion

•Position & Displacement

•Speed & Velocity

•Acceleration

•Newton’s Second Law

•Relative Velocity

3

§3.1 Position & Displacement

The position (r) of an object describes its location relative to some origin or other reference point.

The displacement is the change in an object’s position. It depends only on the beginning and ending positions.

if rrr

4

Example (text problem 3.4): Margaret walks to the store using the following path: 0.500 miles west, 0.200 miles north, 0.300 miles east. What is her total displacement? Give the magnitude and direction.

x

y

r3

r2

r1

r

Take north to be in the +y direction and east to be along +x.

5

Example continued:

The displacement is r = rf ri. The initial position is the origin; what is rf?

The final position will be rf = r1 + r2 + r3. The components are rfx = r1 + r3 = 0.2 miles and rfy = +r2 = +0.2 miles.

miles 283.022 yx rrr

45 and 1tan

x

y

r

r

Using the figure, the magnitude and direction of the displacement are

x

y

rry

rx

N of W.

6

§3.2 Velocity

Velocity is a vector that measures how fast and in what direction something moves.

Speed is the magnitude of the velocity. It is a scalar.

7

Path of a particle

Start

finish

r

vav is the constant speed that results in the same displacement in a given time interval.

tripof time

traveleddistancespeed Average

t

r

vav velocityAverage

t

xv x,av :be wouldcomponent - xThe

8

y

x

ri rf

t

r

vav Points in the direction of r

r

vi

The instantaneous velocity points tangent to the path.vf

A particle moves along the blue path as shown. At time t1 its position is ri and at time t2 its position is rf.

9

On a graph of position versus time, the average velocity is represented by the slope of a chord.

x (m)

t (sec)t1 t2

x1

x2

12

12,av velocityAverage

tt

xxv x

10

tt

rv

0lim velocityousInstantane

This is represented by the slope of a line tangent to the curve on the graph of an object’s position versus time.

x (m)

t (sec)

11

The area under a velocity versus time graph (between the curve and the time axis) gives the displacement in a given interval of time.

v(m/s)

t (sec)

12

Example: Consider Margaret’s walk to the store in the example on slides 3 and 4. If the first leg of her walk takes 10 minutes, the second takes 8 minutes, and the third 7 minutes, compute her average velocity and average speed during each leg and for the overall trip.

t

r

vav velocityAverage

tripof time

traveleddistancespeed Average

Use the definitions:

13

Leg

t

(hours)

vav

(miles/hour)

Average speed

(miles/hour)

1 0.167 3.00 (west) 3.00

2 0.133 1.50 (north) 1.50

3 0.117 2.56 (east) 2.56

Total trip

0.417 0.679(45 N of W)

2.40

Example continued:

14

Example (text problem 3.24): Speedometer readings are obtained and graphed as a car comes to a stop along a straight-line path. How far does the car move between t = 0 and t = 16 seconds?

Since there is not a reversal of direction, the area between the curve and the time axis will represent the distance traveled.

15

Example continued:

The rectangular portion has an area of Lw = (20 m/s)(4 s) = 80 m.

The triangular portion has an area of ½bh = ½(8 s) (20 m/s) = 80 m.

Thus, the total area is 160 m. This is the distance traveled by the car.

16

§3.3 Newton’s Second Law of Motion

t

v

aavonaccelerati Average

A nonzero acceleration changes an object’s state of motion.

tt

va

0limonaccelerati ousInstantane

These have interpretations similar to vav and v.

17

y

x

vi

ri rf

vf

A particle moves along the blue path as shown. At time t1 its position is ri and at time t2 its position is rf.

v

Points in the direction of v.t

vaav

The instantaneous acceleration can point in any direction.

18

Example (text problem 3.39): If a car traveling at 28 m/s is brought to a full stop 4.0 s after the brakes are applied, find the average acceleration during braking.

2av m/s 0.7

s 0.4

m/s 280

t

va

Given: vi = +28 m/s, vf = 0 m/s, and t = 4.0 s.

19

Example (text problem 3.47): At the beginning of a 3 hour trip you are traveling due north at 192 km/hour. At the end, you are traveling 240 km/hour at 45 west of north.

(a) Draw the initial and final velocity vectors.

x (east)

y (north)

vivf

20

(b) Find v.

km/hr 3.2245cos

km/hr 170045sin

ifiyfyy

fixfxx

vvvvv

vvvv

The components are

km/hr 17122 yx vvv

5.71312.0tan1312.0tan 1

x

y

v

vSouth of west

Example continued:

21

(c) What is aav during the trip?

t

v

aav

2av,

2av,

km/hr 43.7hr 3

km/hr 3.22

km/hr 7.56hr 3

km/hr 170

t

va

t

va

yy

xx

The magnitude and direction are:

5.7)1310.0(tan1310.0tan

km/hr 2.57

1

,av

,av

22av,

2av,av

x

y

yx

a

a

aaa

South of west

Example continued:

22

Newton’s 2nd Law:

The acceleration of a body is directly proportional to the net force acting on the body and inversely proportional to the body’s mass.

Mathematically: aFF

a mm

netnet or

23

An object’s mass is a measure of its inertia. The more mass, the more force is required to obtain a given acceleration.

The net force is just the vector sum of all of the forces acting on the body, often written as F.

24

If a = 0, then F = 0. This body can have:

Speed = 0 which is called static equilibrium, or

speed 0, but constant, which is called dynamic equilibrium.

25

§3.4 Applying Newton’s Second Law

aF m

Force units: 1 N = 1 kg m/s2.

26

Example: Find the tension in the cord connecting the two blocks as shown. A force of 10.0 N is applied to the right on block 1. Assume a frictionless surface. The masses are m1

= 3.00 kg and m2 = 1.00 kg.

F block 2 block 1

Assume that the rope stays taut so that both blocks have the same acceleration.

27

FBD for block 2:

TF

w1

N1

x

y

x

T

w2

N2

y

FBD for block 1:

011

1

wNF

amTFF

y

x

022

2

wNF

amTF

y

x

Apply Newton’s 2nd Law to each block:

28

amTF 1

amT 2

These two equations contain the unknowns: a and T.

To solve for T, a must be eliminated. Solve for a in (2) and substitute in (1).

(1)

(2)

N 5.2

kg 1kg 3

1

N 10

1

1

2

1

2

1

21

211

mm

FT

Tm

mT

m

TmF

m

TmamTF

Example continued:

29

Example: A box slides across a rough surface. If the coefficient of kinetic friction is 0.3, what is the acceleration of the box?

Fk

w

N

x

y

FBD for box:

0

k

wNF

maFF

y

x

Apply Newton’s 2nd Law:

30

mgwNwN

maF

0k

mamgNF kkk

(1)

(2)

From (1):

22k m/s 94.2m/s 8.93.0 ga Solving for a:

Example continued:

31

§3.5 Relative Velocity

Example: You are traveling in a car (A) at 60 miles/hour east on a long straight road. The car (B) next to you is traveling at 65 miles/hour east. What is the speed of car B relative to car A?

32east miles/hour 5

eastmiles/hr 60 -east miles/hr 65BA

AGBGBA

AGBGBA

BAAGBG

v

vvv

rrr

rrrFrom the picture:

A

B

A

B

t=0 t>0

rAG

rBG

rBA

Divide by t:

+x

Example continued:

33

Example: You are traveling in a car (A) at 60 miles/hour east on a long straight road. The car (B) next to you is traveling at 65 miles/hour west. What is the speed of car B relative to car A?

34

A

B

A

B

t=0

rAG

rBG rBA

t>0t>0

estmiles/hr w 125

eastmiles/hr 60-estmiles/hr w 65AGBGBA

AGBGBA

vvv

rrr

+x

From the picture:

Divide by t:

Example continued:

35

Summary

•Position

•Displacement Versus Distance

•Velocity Versus Speed

•Acceleration

•Instantaneous Values Versus Average Values

•Newton’s Second Law

•Relative Velocity

36

A ball incident on a wall has a speed of 10 m/s toward the wall. It rebounds with a speed of 10 m/s. What is the direction of the ball’s acceleration while it is in contact with the wall?

A. Toward the wall

B. Away from the wall

C.Up the wall

D.Down the wall

E. The ball is not accelerated.

Additional clicker question: