Summer Assignment Solutions

1.1. Solve:

1.7. Solve: The particle starts with an initial velocity but as it slides it moves slower and slower till coming to rest. This is a case of

negative acceleration because it is an acceleration opposite to the positive direction of motion.

1.11. Solve:

(a)

(b)

1.15. Model: Represent the tile as a particle.

Visualize: The tile falls from the roof with an acceleration equal to a g 9.8 m/s2. Starting from rest, its velocity increases until the tile hits

the water surface. This part of the motion is represented by dots with increasing separation, indicating increasing average velocity. After the

tile enters the water, it settles to the bottom at roughly constant speed.

1.21. Visualize: The bicycle is moving with an acceleration of 1.5 m/s2. Thus, the velocity will increase by 1.5 m/s each second of

motion.

1.25. Solve: (a) 33600 s1 hour 1(hour) 3600 s 3.60 10 s

1 hour

(b) 424 hours 3600 s1 day 1 (day) 8.64 10 s

1 day 1 hour

(c) 4

7365.25 days 8.64 10 s1 year 1 (year) 3.16 10 s

1 year 1 day

(d) 2 2

2

ft 12 inch 1 m32 ft /s 32 9.75 m/s

s 1 ft 39.37 inch

1.29. Solve: (a) 2 3(33.3) 1.109 10 . For numbers starting with “1” an extra digit is kept.

(b) 333.3 45.1 1.50 10

Scientific notation is an easy way to establish significance.

(c) 22.2 1.2 3.5

(d) 1/44.4 0.0225

1.33. Solve: My barber trims about an inch of hair when I visit him every month for a haircut. The rate of hair growth is 2

9

69

1(inch) 2.54 cm 10 m 1 month 1 day 1 h9.8 10 m/s

(month) 1 inch 1 cm 30 days 24 h 3600 s

m 10 m 3600 s9.8 10 35 m/h

s 1 m 1 h

1.37. Model: Represent the speed skater as a particle for the motion diagram.

Visualize:

1.41. Model: Represent the motorist as a particle for the motion diagram.

Visualize:

1.51. Solve:

(a)

(b) Jeremy has perfected the art of steady acceleration and deceleration. From a speed of 60 mph he brakes his car to rest in 10 seconds with a

constant deceleration. Then he turns into an adjoining street. Starting from rest, Jeremy accelerates with exactly the same magnitude as his

earlier deceleration and reaches the same speed of 60 mph over the same distance in exactly the same time. Find the car’s acceleration or

deceleration.

(c)

1.55. Model: The car is represented by the particle model as a dot.

Solve: (a)

Time t (s) Position x (m)

0 1200

10 975

20 825

30 750

40 700

50 650

60 600

70 500

80 300

90 0

(b)

2.1. Model: We will consider the car to be a particle that occupies a single point in space.

Visualize:

Solve: Since the velocity is constant, we have f i .xx x v t Using the above values, we get

1 0 m (10 m/s)(45 s) 450 mx

Assess: 10 m/s 22 mph and implies a speed of 0.4 miles per minute. A displacement of 450 m in 45 s is reasonable and expected.

2.9. Visualize: Please refer to Figure EX2.9.

Solve: (a) The acceleration of the train at 3.0 st is the slope of the v vs t graph at 3 s.t Thus 22 m/s ( 2 m/s) 8 s 0.5 m/sa .

(b)

2.17. Model: We represent the ball as a particle.

Visualize:

Solve: Once the ball leaves the student’s hand, the ball undergoes free fall and its acceleration is equal to the acceleration due to gravity

that always acts vertically downward toward the center of the earth. According to the constant-acceleration kinematic equations of motion

2

1 0 0

1

2y y v t a t

Substituting the known values 2 2

1 12 m 0 m (15 m/s) (1/2)( 9.8 m/s )t t

The solution of this quadratic equation gives 1 3.2 s.t The other root of this equation yields a negative value for

1,t which is not valid for

this problem.

Assess: A time of 3.2 s is reasonable.

2.25. Solve: (a)

(b) To be completed by student.

(c) 22 4 (at 1 s) [2 m/s (1 s) 4 m/s] 2 m/sx x

dxv t v t

dt

(d) There is a turning point at 2 s.t

(e) Using the equation in part (c),

4 m/s (2 4) m/s 4 xv t t

Since 2( 4 2) m, 2 m.x t t x

(f)

2.33. Solve: The position is the integral of the velocity.

1 1 1

0

2 3 31 11 0 0 0 0 13 300

t t t

xt

x x v dt x kt dt x kt x kt

We’re given that 0 9.0 mx and that the particle is at

1 9.0 mx at 1 3.0 s.t Thus

3 31

39.0 m 9.0 m (3.0 s) 9.0 m 9.0 sk k

Solving for k gives 32.0 m/s .k

2.41. Model: The plane is a particle and the constant-acceleration kinematic equations hold.

Solve: (a) To convert 80 m/s to mph, we calculate 80 m/s 1 mi/1609 m 3600 s/h 179 mph.

(b) Using /sa v t , we have,

2 2

s

23 m/s 0 m/s 69 m/s 46 m/s( 0 to 10 s) 2.3 m/s ( 20 s to 30 s) 2.3 m/s

10 s 0 s 30 s 20 ssa t t a t t

For all time intervals a is 2.3 m/s2.

(c) Using kinematics as follows:

2

fs is f i f f( ) 80 m/s 0 m/s (2.3 m/s )( 0 s) 35 sv v a t t t t

(d) Using the above values, we calculate the takeoff distance as follows:

2 2 2

f i is f i s f i

1 1( ) ( ) 0 m (0 m/s)(35 s) (2.3 m/s )(35 s) 1410 m

2 2s s v t t a t t

For safety, the runway should be 3 1410 m 4230 m or 2.6 mi. This is longer than the 2.4 mi long runway, so the takeoff is not safe.

2.49. Model: The rocket is represented as a particle.

Visualize:

Solve: (a) There are three parts to the motion. Both the second and third parts of the motion are free fall, with .a g The maximum

altitude is y2.. In the acceleration phase:

2 2 2 2

1 0 0 1 0 1 0 1

2

1 0 1 0 1

1 1 1( ) ( ) (30 m/s )(30 s) 13,500 m

2 2 2

( ) (30 m/s )(30 s) 900 m/s

y y v t t a t t at

v v a t t at

In the coasting phase,

2 22 2 12 1 2 1 2 1 2

(900 m/s)0 2 ( ) 13,500 m 54,800 m 54.8 km

2 2(9.8 m/s )

vv v g y y y y

g

The maximum altitude is 54.8 km ( 33 miles).

(b) The rocket is in the air until time 3.t We already know

1 30 s.t We can find 2t as follows:

12 1 2 1 2 10 m/s ( ) 122 s

vv v g t t t t

g

Then t3 is found by considering the time needed to fall 54,800 m:

2 2 23 2 2 3 2 3 2 2 3 2 3 2

1 1 20 m ( ) ( ) ( ) 228 s

2 2

yy y v t t g t t y g t t t t

g

(c) The velocity increases linearly, with a slope of 30 (m/s)/s, for 30 s to a maximum speed of 900 m/s. It then begins to decrease linearly

with a slope of 9.8(m/s)/s. The velocity passes through zero (the turning point at 2y ) at

2 122 s.t The impact velocity at 3 228 st is

calculated to be 3 2 3 2( ) 1040 m/s.v v g t t

Assess: In reality, friction due to air resistance would prevent the rocket from reaching such high speeds as it falls, and the acceleration

upward would not be constant because the mass changes as the fuel is burned, but that is a more complicated problem.

2.57. Model: We will use the particle model for the puck.

Visualize:

We can view this problem as two one-dimensional motion problems. The horizontal segments do not affect the motion because the speed

does not change. So, the problem “starts” at the bottom of the uphill ramp and “ends” at the bottom of the downhill ramp. At the top of the

ramp the speed does not change along the horizontal section. The final speed from the uphill roll (first problem) becomes the initial speed of

the downhill roll (second problem). Because the axes point in different directions, we can avoid possible confusion by calling the downhill

axis the z-axis and the downhill velocities u. The uphill axis as usual will be denoted by x and the uphill velocities as v. Note that the height

information, h 1 m, has to be transformed into information about positions along the two axes.

Solve: (a) The uphill roll has 2

0 sin30 4.90 m/s .a g The speed at the top is found from

2 2

1 0 0 1 02 ( )v v a x x

2 2 2

1 0 0 12 (5 m/s) 2( 4.90 m/s )(2.00 m) 2.32 m/sv v a x

(b) The downward roll starts with velocity 1 1 2.32 m/su v and 2

1 sin20 3.35 m/s .a g Then,

2 2 2 2

2 1 1 2 1 22 ( ) (2.32 m/s) 2(3.35 m/s )(2.92 m 0 m) 5.00 m/su u a z z u

(c) The final speed is equal to the initial speed, so the percentage change is zero!

Assess: This result may seem surprising, but can be more easily understood after we introduce the concept of energy. For now, imagine this

is a one dimensional vertical problem. The total vertical change in height of the puck is zero. We have already seen how an object with an

initial velocity upward has the same velocity in the opposite direction as it passes through that height going down.

2.65. Solve: (a) The quantity4

2 32 2(3.6 10 W)60 m /s .

1200 kg

P

m

Thus

2

3m60

sxv t

At 2

3m10 s, 60 (10 s) 24 m/s ( 50 mph),

sxt v and at t 20 s, 35 m/sxv ( 75 mph).

(b) With1

22

, we havex

Pv t

m

12

2 1

2 2

xx

dv P Pa t

dt m mt

(c) At 1 s,t 4

2(3.6 10 W)= 3.9 m/s .

2 2(1200 kg)(1 s)x

Pa

mt

Similarly, at 210 s, 1.2 m/s .xt a

(d) Consider the limiting case of very short times. Note that as 0.xa t This is physically impossible for the Alfa Romeo.

(e) We can use the relationship that x

dxv

dt and integrate to find ( ).x t We have

12

2x

Pv t

m and the initial condition 0ix at 0.it Thus

1/ 2

0 0

2x tPdx t dt

m

3/ 23/ 22 2 2

and 3/ 2 3

P t Px t

m m

(f) Time to travel a distance x is found by solving the above equation for t.

2 / 3

3

2 2

mt x

P

For 402 m, 18.2 s.x t

2.73. Solve: (a) A comparison of this equation with the constant-acceleration kinematic equation

2 2

1y 0y y 1 0( ) 2( )( )v v a y y

yields the following information: 2

0 10 m, 10 m, 9.8 m/s ,yy y a and 1 10 m/s.yv It is clearly a problem of free fall. On a romantic

Valentine’s Day, John decided to surprise his girlfriend, Judy, in a special way. As he reached her apartment building, he found her sitting in

the balcony of her second floor apartment 10 m above the first floor. John quietly armed his spring-loaded gun with a rose, and launched it

straight up to catch her attention. Judy noticed that the flower flew past her at a speed of 10 m/s. Judy is refusing to kiss John until he tells her

the initial speed of the rose as it was released by the spring-loaded gun. Can you help John on this Valentine’s Day?

(b)

(c) 2 2 2

0y 0y(10 m/s) 2(9.8 m/s )(10 m 0 m) 17.2 m/sv v

Assess: The initial velocity of 17.2 m/s, compared to a velocity of 10 m/s at a height of 10 m, is very reasonable.

2.81. Model: We will use the particle-model to represent the sprinter and the equations of kinematics.

Visualize:

Solve: Substituting into the constant-acceleration kinematic equations,

2 2 2 2

1 0 0 1 0 0 1 0 0 0 1 0

2

1 0

1 0 0 1 0 0 1 0

1 1 1 1( ) ( ) 0 m 0 m (4 s 0 s) (4.0 s)

2 2 2 2

(8 s )

( ) 0 m/s (4.0 s 0 s) (4.0 s)

x x v t t a t t a a t a

x a

v v a t t a v a

From these two results, we find that 1 1(2 s) .x v Now,

2

2 1 1 2 1 1 2 1

1 1 1

1( ) ( )

2

100 m (2 s) (10 s 4 s) 0 m 12.5 m/s

x x v t t a t t

v v v

Assess: Using the conversion 2.24 mph 1 m/s, 1 12.5 m/s 28 mph.v This speed as the sprinter reaches the finish line is physically

reasonable.

3.1. Visualize:

Solve: (a) To find A B , we place the tail of vector B on the tip of vector A and connect the tail of vector A with the tip of vector .B

(b) Since ( )A B A B , we place the tail of the vector ( )B on the tip of vector A and then connect the tail of vector A with the tip of

vector ( )B .

3.5. Visualize: The figure shows the components vx and vy, and the angle .

Solve: We have, sin40 ,yv v or 10 m/s sin40 ,v or 15.56 m/s.v

Thus the x-component is cos40 (15.56 m/s ) cos40 12 m/s.xv v

Assess: The x-component is positive since the position vector is in the fourth quadrant.

3.9. Visualize:

Solve: The magnitude of the vector is2 2 2 2( ) ( ) (125 V/m) ( 250 V/m) 280 V/m.x yE E E In the expression for ,E the j and

i means that E is in quadrant IV. The angle is below the positive x-axis. We have:

1 1 1| | 250 V/m

tan tan tan 2 63.4125 V/m

y

x

E

E

Assess: Since | | | |y xE E , the angle made with the x-axis is larger than 45. 45 for | | | |y xE E .

3.13. Visualize: The vectors , ,A B and C A B are shown.

Solve: (a) We have ˆ ˆ5 2A i j and ˆ ˆ3 5 .B i j Thus, C A B ˆ ˆ ˆ ˆ(5 2 ) ( 3 5 )i j i j ˆ ˆ2 3 .i j

(b) Vectors ,A ,B and C are shown with their tails together.

(c) Since ˆ ˆ ˆ ˆ2 3 ,x yC i j C i C j 2,xC and 3.yC Therefore, the magnitude and direction of C are 2 2(2) ( 3) 3.6C

1 1| | 3

tan tan 56 below the -axis+2

y

x

Cx

C

Assess: The vector C is to the right and down, thus implying a negative y-component and positive x-component, as obtained above. Also

45 since | | | |y xC C .

3.17. Solve: A different coordinate system can only mean a different orientation of the grid and a different origin of the grid.

(a) False, because the size of a vector is fixed.

(b) False, because the direction of a vector in space is independent of any coordinate system.

(c) True, because the orientation of the vector relative to the axes can be different.

3.21. Visualize:

Solve: Using the method of tail-to-tip graphical addition, the diagram shows the resultant for D E F in (a), the resultant for 2D E in

(b), and the resultant for 2D E F in (c).

3.29. Visualize:

The magnitude of the unknown vector is 1 and its direction is along ˆ ˆi j . Let ˆ ˆA i j as shown in the diagram. That is, ˆ ˆ1 1A i j and

the x- and y-components of A are both unity. Since 1tan ( / ) 45 ,y xA A the unknown vector must make an angle of 45 with the x-axis

and have unit magnitude.

Solve: Let the unknown vector be ˆ ˆx yB B i B j where

1 1cos45 and sin45

2 2x yB B B B B B

We want the magnitude of B to be 1, so we have

2 2

2 2

2

1

1 11 1 1

2 2

x yB B B

B B B B

Hence,

1

2x yB B

Finally,

ˆ ˆx yB B i B j

1 1ˆ ˆ2 2

i j

3.33. Visualize: (a)

Note that x is along the east and y is along the north.

Solve: (b) We are given ˆ(200 m) ,A j and can use trigonometry to obtain ˆ ˆ(283 m) (283 m)B i j and ˆ ˆ(100 m) (173 m) .C i j We

want 0.A B C D This means

ˆ ˆ ˆ ˆ ˆ ˆ ˆ(200 m ) (283 m 283 m ) ( 100 m 173 m ) 183 m 310 m

D A B C

j i j i j i j

The magnitude and direction of D are

2 2 1 1 310 m(183 m) (310 m) 360 m and tan tan 59.4

183 m

y

x

DD

D

This means (360 m, 59.4D north of east).

(c) The measured length of the vector D on the graph (with a ruler) is approximately 1.75 times the measured length of vector A . Since A

200 m, this gives D 1.75 200 m 350 m. Similarly, the angle measured with the protractor is close to 60. These answers are in close

agreement to part (b).

3.37. Visualize:

Solve: (a) Since cos ,xv v we have 2.5 m/s (3.0 m/s)cos 1 2.5 m/scos 34 .

3.0 m/s

(b) The vertical component is sinyv v (3.0 m/s) sin34 1.7 m/s.

3.41. Visualize: A 3% grade rises 3 m for every 100 m horizontal distance. The angle of the ground is thus1 1tan (3/100) tan (0.03) 1.72 .

Establish a tilted coordinate system with one axis parallel to the ground and the other axis perpendicular to the ground.

Solve: From the figure, the magnitude of the component vector of v perpendicular to the ground is sin 15.0 m/s.v v

But this is only the size. We also have to note that the direction of v is down, so the component is v

15.0 m/s.

3.45. Visualize:

Use a tilted coordinate system such that x-axis is down the slope.

Solve: Expressing all three forces in terms of unit vectors, we have 1ˆ(3.0 N) ,F i 2

ˆ(6.0 N) ,F j and

3F ˆ ˆ(5.0 N)sin (5.0 N)cos .i j

(a) The component of netF parallel to the floor is net( ) (3.0 N) 0 N (5.0 N)sin30 0.50 N,xF or 0.50 N up the slope.

(b) The component of netF perpendicular to the floor is net( ) 0 N (6.0 N) (5.0 N)cos30 1.67 N.yF

(c) The magnitude of netF is net net net( ) ( )x yF F F 2 2( 0.50 N) (1.67 N) 1.74 N. The angle netF makes is

net1 1

net

( ) 1.67 Ntan tan 73

|( ) | 0.50 N

y

x

F

F

netF is 73 above the floor on the left side of 2.F 00000