1

Chapter 2Unit Vector

Vector addition by components

Vector addition by graphing

2

Displacement (Vector)

0xxx

xxx 0

0ttt

3

Speed & Velocity

Average and Instantaneous Average Speed

= Distance/ Elapsed time

(How fast it moves, but no direction)

Units and dimensions

4

0

0

tt

xx

t

xv

5.339074.4

01609

7.3420695.4

01609

a)m/s

b)

m/s

Time to go to Florida from Atlanta Average Velocity?

Average Velocity

5

Instantaneous Velocity

Smaller and smaller gives smaller and smaller As 0 it gives the limiting value

(In general instantaneous values are given unless noted otherwise.)

What if changes ( push the accelerator pedal)

t

xLimitv

t

0

t x

t

v

6

Acceleration

Rate of change of velocity is acceleration

Average accelerationt

v

tt

vva

0

0

7

Equation of KinematicsConstant acceleration (straight line- motion)

Say X0 and t0 = 0

t

vvaa 0

atvv 0

t

x

tt

xxv

0

0 0,0 00 tx

atvv 0

tvx

8

Constant acceleration

Velocity increases

What is the average velocity?

20 vv

v

tvv

tvx

2

0

tvvx 02

1

9

tvvx 02

1 tatvv 002

1

tattv 2

12

2

10

If starts at 0,0 tx

and with constant acceleration.

20 2

1attvx

10

If you do not know “t” tvvx 02

1

atvv 0

ta

vv

0

a

vvvvx 0

02

1

a

vv 20

2

2

1

axvv 220

2 axvv 22

02

11

Summary

Equation Variables

number Equation x a v v0 t(2.4) v=v0+at -----

(2.7) x=(1/2)(v0+v)t -----

(2.8) x=v0t+(1/2)at2 -----

(2.9) v2=v02+2ax -----

Table 2.1 Equations of Kinematics for Constant Acceleration

12

Application of the Equations

Draw the situation Pick +ve and –ve direction and co-

ordinates Write the known values (with signs) Look for hidden information Enough information, what is unknown? If two objects, are they connected?

13

Free FALL No air resistance Constant acceleration due to gravity

g= 9.80 m/s2 or 32.2 ft/s2 (on earth surface)

Gallileo’s test Leaning tower (better on moon)

Air filled tube Evacuated tube

14

Example 10 A Falling StoneAfter 3.0 sec what is the displacement

v0=0 m/s

t=3.0 sec

Known: v0=0 a=g (i.e.)

t=3.0

y=v0t+(1/2)at2

v=v0+at

v2=v02+2ay

y=v0t+(1/2)at2

=0+(1/2)(-9.8)*9=(-44.1m)

Which eqn to use?

15

What is the velocity after 3.0 sec?

v=v0+at

v2=v02+2ay

Can use both----why?

Which is easier?

v = -9.8m/s2*3sec = -29.4ms

-ve sign indicate

16

Example 12 How high the coin reach, if it is tossed up with an initial speed of 5.00 m/s?

At the highest point the velocity should be zero.

v0=5.0 m/s, a=-9.8 m/s2

v=0.0 m/s, y=?

v2=v02+2ay

ma

vvy 28.1

8.92

25

8.92

50

2

220

2

17

How long will it take to reach the top ?

v=0, v0=5, a= -9.8

How much time it takes to come to the same level as it was thrown ?

v0=5.0

a=-9.8

y=0 t=?

Which eqn ?

v=v0+at 0=5-9.8t t=0.5/sec

18

20 2

1attvy

28.92

150 tt tt

2

8.95

t=0 or 02

8.95 t

i.e.t=0 or sec02.18.9

52 t

Why two values? Explain each answer.

19

symmetryNo air resistance—acceleration is the same for the

motion in both directions

Velocity vector changes continuously

Acceleration does not.

Time to reach the highest point

= time to reach the ground

At same height, speed is the same.

20

Bullet reaches the ground with the same speed irrespective of which direction it was fired first.

21

150 m

Lets say

Show the speed of the bullet when it comes down to the same height.

What is the velocity if shot up with initial velocity 30 m/s under gravity when it reaches the same level?

Known information: v0=30, y=0,

Which eqn ?

a=-g

v2=v02+2ay

v2=(30)2+2(-9.8)*0

v=+/-30

What is the significance of + and – sign?

22

Harder way

How high it reaches before changing direction?

v0= 30, a=-g, v=?(hidden information) y=?(needs to find)

Which eqn ?

v2=v02+2ay

0=(30)2+2(-9.8)*y

y=(30*30)/(2*9.8)=45.92 m

23

v0=0

v=?

y=?

For the downward motion,

v2=v02+2ay

=0+2(+9.8)*[(30*30)/(2*9.8)] (why +ve ?) =30*30v=30 m/s (again +/-)

Find the time to reach max height and then find the speed when it reaches the same level.

24

What is the velocity of the bullet when it reaches the ground if we start from the mountain?

v=? v0= 30, a=-9.8, y=-150 (why –ve ?)

v2=v02+2ay

v2=(30)2+2(-9.8)(-150) =(30)2+2.94*103

=3.84*103

v=61.97 m/s (which sign?)

25

If you drop down with 30 m/s from the mountain, what is the speed?

v0= -30, g=-9.8, y=-150

v2=v02+2ay

=(30)2+2(-9.8)(-150) =(30)2+2.94*103

=3.84*103

v=61.97 m/s

26

Example

t=0 v0

t=t

2v0

v=v0+at

0=v0+at

v=2v0+at1

0=2v0+at1

8.90

v

t tv

t 28.9

2 01

27

ayvv 220

2

yv 8.920 20

8.92

20

vy

12

02 22 ayvv

12

0 8.9220 yv

yv

y 48.92

4 20

1

Answer-- 2t, 4y

28

Graphical analysis

t

xv

x=0 and t=0

Slope = smt

x/4

29

Example 16 Bicycle Trip

30

smt

xv /2

200400

400800

smt

xv /0

6001000

0

smt

xv /1

14001800

800400

1st segment

2nd segment

3rd segment

31

If velocity changes

(i.e. acceleration)

20 2

1attvx

(assume v0=0 for simplicity)

Slope of the tangent

Instantaneous velocity

v=v0+at

32

Conceptual questions 2

REASONING AND SOLUTION The buses do not have equal velocities. Velocity is a vector, with both magnitude and direction. In order for two vectors to be equal, they must have the same magnitude and the same direction. The direction of the velocity of each bus points in the direction of motion of the bus. Thus, the directions of the velocities of the buses are different. Therefore, the velocities are not equal, even though the speeds are the same.

33

Conceptual question 4

REASONING AND SOLUTION Consider the four traffic lights 1, 2, 3 and 4 shown below. Let the distance between lights 1 and 2 be x12, the distance between lights 2 and 3 be x23, and the distance between lights 3 and 4 be x34.

v

x12 x34x23

1 2 3 4

34

The lights can be timed so that if a car travels with a constant speed v, red lights can be avoided in the following way. Suppose that at time t = 0 s, light 1 turns green while the rest are red. Light 2 must then turn green in a time t12, where t12 = x12/v. Light 3 must turn green in a time t23 after light 2 turns green, where t23 = x23/v. Likewise, light 4 must turn green in a time t34 after light 3 turns green, where t34 = x34/v. Note that the timing of traffic lights is more complicated than indicated here when groups of cars are stopped at light 1. Then the acceleration of the cars, the reaction time of the drivers, and other factors must be considered.

35

Conceptual question 5

REASONING AND SOLUTION The velocity of the car is a vector quantity with both magnitude and direction The speed of the car is a scalar quantity and has nothing to do with direction. It is possible for a car to drive around a track at constant speed. As the car drives around the track, however, the car must change direction. Therefore, the direction of the velocity changes, and the velocity cannot be constant. The incorrect statement is (a).

36

Conceptual question 14REASONING AND SOLUTION The magnitude of the muzzle velocity of the bullet can be found (to a very good approximation) by solving Equation 2.9,

2 20 2v v a x With v0 = 0 m/s; that is

v 2ax

where a is the acceleration of the bullet and x is the distance traveled by the bullet before it leaves the barrel of the gun (i.e., the length of the barrel).

37

Since the muzzle velocity of the rifle with the shorter barrel is greater than the muzzle velocity of the rifle with the longer barrel, the product ax must be greater for the bullet in the rifle with the shorter barrel. But x is smaller for the rifle with the shorter barrel, thus the acceleration of the bullet must be larger in the rifle with the shorter barrel.

38

Problem 4

3 37.6 10 m / s 110 10 sNumber = 9.1

91.4 m

v tx

L L

REASONING The distance traveled by the Space Shuttle is equal to its speed multiplied by the time. The number of football fields is equal to this distance divided by the length L of one football field.

SOLUTION The number of football fields is

39

Problem 8REASONING AND SOLUTION Let west be the positive direction. The average velocity of the backpacker is

w e w ewhere andw e

w e w e

x x x xv t t

t t v v

.

40

Combining these equations and solving for xe (suppressing the units) gives

The distance traveled is the magnitude of xe , or 0.81 km

41

Problem 8 solution (2)N

S

EW

y

xO

N

S

EW

y

xO

6.44 km

A

Time for 1st segmentsm

mt

/68.2

64401

O A is 6.44 km with an velocity of 2.68 m/s =say t1

N

S

EW

y

xO

6.44 km

A

xB

(X in meters)

A B is x m with an velocity of .447m/s =say t2

Time for 2nd segment sec447.2

xt

Time for displacement(O B) with an average velocity of 1.34m/s =say t

sec34.1

6440 xt

42

t1+t2=t

34.1

6440

447.68.2

6440 xx

34.1

6440

447.99.2402

xx

447.0)6440()34.1(34.1447.99.2402 xx

xx 447.0108784.234.11043934.1 33

3104393.1787.1 x

kmmx 81.0805

43

Problem 10Reasoning The definition of average velocity is given by Equation 2.2 as Average velocity = Displacement/(Elapsed time). The displacement in this expression is the total displacement, which is the sum of the displacements for each part of the trip. Displacement is a vector quantity, and we must be careful to account for the fact that the displacement in the first part of the trip is north, while the displacement in the second part is south.

44

Solution According to Equation 2.2, the displacement for each part of the trip is the average velocity for that part times the corresponding elapsed time. Designating north as the positive direction, we find for the total displacement that

+27 m/s for tn

xn

-17 m/sxs

North South

Northward Southward

Displacement = 27 m/s 17 m/st t

Displacement = xn+ xs

45

Where tNorth and tSouth denote, respectively, the times for each part of the trip. Note that the minus sign indicates a direction due south. Noting that the total elapsed time is tNorth + tSouth we can use Equation 2.2 to find the average velocity for the entire trip as follows:

North South

North South

North South

North South North South

27 m/s 17 m/sDisplacementAverage velocity =

Elapsed time

= 27 m/s 17 m/s

t t

t t

t t

t t t t

46

North South

North South North South

3 1 and

4 4

t t

t t t t

But

Therefore, we have that

3 1Average velocity = 27 m/s 17 m/s 16 m/s

4 4

The plus sign indicates that the average velocity for the entire trip points north.

47

Problem 19REASONING AND SOLUTION

x 1

2a t 2

1

2

6.0 m/s

1.5 s

1.5 s 2 4.5 m

Uniform acceleration, starts from rest.

x=v0t+(1/2)at2

48

Problem 23

REASONING We know the initial and final velocities of the blood, as well as its displacement. Therefore, Equation 2.9 can be used to find the acceleration of the blood. The time it takes for the blood to reach it final velocity can be found by using Equation 2.7

102

xt

v v

2 20 2v v ax

49

SOLUTION

a.) The acceleration of the blood is

2 22 22 20

26 cm / s 0 cm / s1.7 10 cm / s

2 2 2.0 cm

v va

x

b.) The time it takes for the blood, starting from 0 cm/s, to reach a final velocity of +26 cm/s is

1 102 2

2.0 cm0.15 s

0 cm / s + 26 cm / s

xt

v v

50

Problem 28REASONING AND SOLUTION The speed of the car at the end of the first (402 m) phase can be obtained as follows: v1

2 = vo2 + 2a1x1

v1 = 2(17.0 m/s2)(402 m)

The speed after the second phase (3.50 102 m) can be obtained in a similar fashion. v2

2 = v022 + 2a2x2

v2 = v

12 + 2(- 6.10 m/s2)(3.50 x 10

2 m)

v2=96.9 m/s

51

Problem 39REASONING The initial velocity and the elapsed time are given in the problem. Since the rock returns to the same place from which it was thrown, its displacement is zero (y = 0 m). Using this information, we can employ Equation 2.8 y=v0t+(1/2)at2 to determine the acceleration a due to gravity.

SOLUTION Solving Equation 2.8 for the acceleration yields

0 22 2

2 0 m 15 m / s 20.0 s2 1.5 m / s

20.0 s

y v ta

t

52

Problem 44REASONING AND SOLUTION

a. v 2 v02 2 ay

v 1.8 m/s 2 2 –9.80 m/s 2 –3.0 m 7.9 m/s

The minus is chosen, since the diver is now moving down. Hence, v=-7.9 m/s.

53

b. The diver's velocity is zero at his highest point. The position of the diver relative to the board is

y –v0

2

2a–

1.8 m/s 2

2 –9.80 m/s 2 0.17 m

The position above the water is 3.0 m + 0.17 m = 3.2 m/s

54

Problem 48t

t0=0 v0=25.0 m/s

t

t0=1.2 sec v0=?

v=v0+at

0=25-9.8t

t=25/9.8=2.55 sec

v=v0+at

0=v0-9.8(t-1.2)

v0= 9.8(t-1.2)= 9.8(2.55-1.2)

=9.8*1.35=13.23 m/s

55

Problem 51v0=0

75 m

5.0 m/sx=?

What is the time for stone to hit the water(log)?

y=v0t+(1/2)at2

(-75)=(1/2)*(-9.8)t2

t2=2*75/9.8=15.3 sec

If the log is moving with constant velocity during this time, the log should have moved.

x=vt=5*3.9=19.5 m

56

Problem 55REASONING AND SOLUTION The balls pass at a time t when both are at a position y above the ground. Applying Equation 2.8 to the ball that is dropped from rest, we have

1 12 201 2 2

= 24 m + + 24 m + 0 m/s +y v t at t at (1)

Note that we have taken into account the fact that y = 24 m when t = 0 s in Equation (1). For the second ball that is thrown straight upward, y = v02t +

1

2at 2

Equating Equations (1) and (2) for y yields

tvattvat 022

2

102

22

1 = m 24or+=+m 42

57

Thus, the two balls pass at a time t, where t 24 m

v02The initial speed v02 of the second ball is exactly the same as that with which the first ball hits the ground. To find the speed with which the first ball hits the ground, we take upward as the positive direction and use Equation 2.9 v2=v0

2+2ay . Since the first ball is dropped from rest, we find that

smmsm /7.2124/8.92 2

58

At a time t = 1.11 s, the position of the first ball according to Equation (1) is

y 24 m +1

2(–9.80 m/s2 )(1.11 s)2 24 m – 6.0 m

which is 6.0m below the top of the cliff.

Thus, the balls pass after a time

t 24 m

21.7 m/s1.11 s