CHAPTER 2 MOTION ALONG A STRAIGHT LINE … · 12 CHAPTER 2 MOTION ALONG A STRAIGHT LINE Solutions to Exercises & Problems 1E ... 6E Calculate the time ... CHAPTER 2 MOTION ALONG A

CHAPTER 2 MOTION ALONG A STRAIGHT LINE 11

CHAPTER 2

Answer to Checkpoint Questions

1. (b) and (c)

2. zero

3. (a) (1) and (4); (b) (2) and (3); (c) (3)

4. (a) plus; (b) minus; (c) minus; (d) plus

5. (1) and (4)

6. (a) plus; (b) minus; (c) a = �g = �9:8m/s2.

Answer to Questions

1. (a) yes; (b) no; (c) yes; (d) yes

2. (a) minus; (b) positive; (c) zero; (d) negative; (e) twice (at t = 1 s and 3 s)

3. (a) 2, 3; (b) 1, 3; (c) 4

4. (a) negative direction of x; (b) positive direction of x; (c) yes (when graphed line

crosses the t axis); (e) positive; (f) constant

5. all tie (see Eq. 2-16)

6. a and c

7. (a) �g; (b) 2m/s upward

9. same

10. 40mi/h, not 0

11. x = t2 and x = 8(t� 2) + (1:5)(t� 2)2

12. 1 and 2 tie, then 3

13. increase

12 CHAPTER 2 MOTION ALONG A STRAIGHT LINE

Solutions to Exercises & Problems

1E

(a) Use �v = �x=�t to �nd the average speed. For Carl Lewis �vL = 100m=10 s = 10m/s;

and for Bill Rodgers

�vR =26mi 395 yd

2 h 10min=

(26mi)(1609m/mi) + (395 yd)(0:9144m/yd)

7800 s= 5:41m/s :

(b) The time it would have taken for Lewis to �nish the marathon is

�t =�x

�vL=

42186m

10m/s= 4219 s = 1h 10min 19 s :

2E

The car has moved a distance of

s = vt = (90 km/h)(0:50 s) = (90 km/h)(1000m/km)(1 h=3600 s)(0:50 s) = 13m :

3E

The distance the �ghter travels in a blink is

�x = �v�t = (2110mi/h)(100ms)

�3600 s/h

1609m/mi

�= 94:3m = 309 ft :

4E

(a) Assume that the ball travels at a constant velocity and use �x = v�t, where �x

is the horizontal distance traveled, �t is the time, and v is the velocity. Convert v to

meters per second. According to Appendix D, 1 km/h = 0:2778m/s, so 160 km/h =

(160)(0:2778m/s) = 44:45m/s. Thus

�t =�x

v=

18:4m

44:5m/s= 0:414 s :

This may also be written 414ms.

x (km)

20

80

1.0t (h)

0.5

40

60

0 1.5 2.0


5E

The average speed can be found from Fig. 2-18 of the text:

�v ��

1800 km

90� 106 y

��105 cm

1 km

�= 2 cm/y :

6E

Calculate the time required to make the trip at each of the two speeds and compute the

di�erence. Use �x = v�t. At the slower speed the time required is

�t =�x

v=

700 km

88:5 km/h= 7:91 h ;

and at the faster one

�t =�x

v=

700 km

105 km/h= 6:67 h :

The di�erence is 7:91 h� 6:67 h = 1:24 h = 1h 14min.

7E

The speed of light c in various units is

c = 3� 108m/s = (3� 108m/s)(6:214� 10�4mi/m)(3600 s=h) = 6:71� 108mi/h

= (3� 108m/s)(3:281 ft/m) = 9:84� 108 ft/s = 1:00 ly/y :

8E

The average velocity during some time interval �t is �v = �x=�t, where �x is the dis-

placement. In this case the interval is divided into two parts. During the �rst part twe

have �x1 = 40 km and �t1 = (40 km)=(30 km/h) = 1:33 h. During the second part we

have �x2 = 40 km and �t2 = (40 km)=(60 km/h) = 0:67 h. Both displacements are in the

same direction so the total displacement is �x = �x1 + �x2 = 40 + 40 = 80 km. The

total time interval is �t = �t1 + �t2 = 1:33 h + 0:67 h = 2:00 h. The average velocity is

�v = (80 km)=(2:0 h) = 40 km/h.

(b) The average speed is the total distance

traveled divided by the time. In this case

the total distance is the magnitude of the

total displacement, so the average speed is

40 km/h.

(c) Assume the automobile starts at the ori-

gin at time t = 0. Then its coordinate as a

function of time is as shown as the solid

480

240

x (ft)

t(s)60 84

slope=v v2

v1

840

240

x (ft)

60 120

slope=v

v2

v1

0 0


lines on the graph in the last page. The average velocity is the slope of the dotted line.

9P

(a)

�v =�x

�t=

(240 + 240) ft

(240=4 + 240=10) s= 5:7 ft/s :

(b)

�v =�x

�t=

[(60)(4) + (60)(10)] ft

(60 + 60) s= 7:0 ft/s :

(c)

10P

Let the length of the hill be D. Then the total time of travel is

t =D

40 km/h+

D

60 km/h:

The average speed is thus given by

�v =2D

t=

2D

D=(40 km/h) +D=(60 km/h)= 48 km/h :

480

240

x (ft)

t(s)60 84

slope=v v2

v1

840

240

x (ft)

60 120

slope=v

v2

v1

0 0


11P

(a) Let the distance between San Antonio and Houston be L, and the time it takes to travel

to Houston be T . Then on the way to Houston, the average speed is

�v1 =L

T=

�T

2

�(35mi/h) +

�T

2

�(55mi/h)

T= 45mi/h :

(b) On the way back from Houston, the average speed is

�v2 =L

L

2(35mi/h)+

L

2(55mi/h)

= 43mi/h :

(c) The average speed for the entire trip is

�v =2L

L=(45mi/h) + L=(43mi/h)= 44mi/h :

(d) Since there is no net displacement in a round trip, the average velocity is zero.

(e)

Refer to the �gure above. The average velocity is given by the slope of the dashed line,

with the positive sign indicating that the motion is in the positive x direction.

12P

(a) Substitute, in turn, t = 1, 2, 3, and 4 s into the expression x(t) = 3t� 4t2 + t3:

x(1 s) = (3)(1)� (4)(1)2 + 13 = 0

x(2 s) = (3)(2)� (4)(2)2 + 23 = �2mx(3 s) = (3)(3)� (4)(3)2 + 33 = 0

x(4 s) = (3)(4)� (4)(4)2 + 43 = 12m :

(b) The displacement during an interval is the coordinate at the end of the interval minus

the coordinate at the beginning. For the interval from t = 0 to t = 4 s, the displacement is

�x = x(4 s)� x(0) = 12m� 0 = +12m. The displacement is in the positive x direction.

x (m)

3.0

12.0

1.0t (s)

6.0

9.0

0.03.02.0 4.0

-3.0


(c) The average velocity during an interval is de�ned as the displacement over the interval

divided by the interval: �v = �x=�t. For the interval from t = 2 s to t = 4 s the displacement

is �x = x(4 s)�x(2 s) = (12m)�(�2m) = 14m and the time interval is �t = 4 s�2 s = 2 s.

Thus

�v =�x

�t=

14m

2 s= 7m/s :

(d) The solid curve on the graph below shows the coordinate x as a function of time. The

slope of the dotted line is the average velocity between t = 2:0 s and t = 4:0 s.

13P

(a) The average velocity during the time interval between t = 2:00 s and t = 3:00 s is

�v2 s!3 s =�x

�t=

x(t = 3:00 s)� x(t = 2:00 s)

3:00 s� 2:00 s

=1:5[(3:00)3 � (2:00)3] cm

1:00 s= 28:5 cm/s :

(b) The instantaneous velocity at time t is given by v(t) = dx=dt = 4:5t2, which gives

v(t = 2:00 s) = (4:5)(2:00)2 cm/s = 18:0 cm/s :

(c)

v(t = 3:00 s) = (4:5)(3:00)2 cm/s = 40:5 cm/s :

(d)

v(t = 2:50 s) = (4:5)(2:50)2 cm/s = 28:1 cm/s :

(e) Suppose that, at time tm, the particle is midway between its positions at t = 2:00 s and

t = 3:00 s. Then

x(tm) = 9:75 + 1:5t3m =x(t = 2:00 s) + x(t = 3:00 s)

2

=[9:75 + 1:5(2:00)3] + [9:75 + 1:5(3:00)3]

2;

x (cm)

t (s)

v(2s)

v(2.5s)

v(3s)vm

20

0

40

60

80

100

120

1.5 2.0 2.5 3.0 3.5

x(t)

vp t

θ


yielding tm = 2:596 s. Thus the instantaneous speed at this instant is v(tm) = 4:5(2:596)2 =

30:3 cm/s:

(f) The answers to the preceeding questions are given by the slopes in the �gure below.

14P

Let the speed of the jet plane be vp. From the �gure above it is obvious that the maximum

time t that the pilot has before a correction has to be made satisi�es h=vpt = tan �, which

gives

t =h

vp tan �=

35m

(1300 km/h)(1000m/km)(1 h=3600 s)(tan 4:3�)= 1:3 s :


15P

(a) Neglect the size of the bird. Let Dn be the separation between the two trains when the

bird begins its nth trip, then

Dn+1 = Dn � 2Dn

�30 km/h

30 km/h + 60 km/h

�=

1

3Dn ;

which gives Dn = (1=3)n�1D1, where D1 = 60 km. The total number of trips N that the

bird makes before the crash thus satis�es

DN =

�1

3

�N�1D1 = 0 ;

which gives N =1 :

(b) The total time elapsed before the two trains crash is

t =60 km

(30 + 30) km/h= 1h ;

during which the bird has traveled a distance of x = vt = (60 km/h)(1 h) = 60 km:

16E

(a) The velocity of the particle is

v(t) =dx

dt=

d

dt(4� 12t+ 3t2) = �12 + 6t :

Thus v(t = 1 s) = [�12 + (6)(1)]m/s = �6m/s.

(b) Since v < 0, it is moving in the negative x direction.

(c) The speed is jv(t = 1 s)j = 6m/s.

(d) For 0 < t < 2 s, jv(t)j decerases with time. At t = 2 s, v(t) is zero. Afterwards, jv(t)jincreases with time.

(e) Yes. Since v crosses over from negative (at t = 0) to positive (at large t), it has to pass

through zero. This actually happens at t = 2 s.

(f) In fact, from v(t) = �12 + 6t, we know that v(t) > 0 for t > 2 s. Thus the particle

cannot move leftward on the x axis after t = 3 s.

17E

The sign of the velocity is determined by whether the local slope of the x vs t curve is

positive or negative.

(a) The animal is to the left of the origin on the axis between t = 2:0 s and t = 4:0 s.

(b) The velocity is negative between t = 0 and t = 3:0 s.

(c) The velocity is positive between t = 3:0 s and t = 7:0 s.

(d) The velocity is zero at t = 3:0 s.

(1)

(2)

(3)

(4)

steepest

least steep

x0 t

v (m/s)

20

1.0t (s)

0.50

10

1.5

-10

-20

2.0 2.5


18E

The x vs t curve is shown below. The steepest part corresponds to the highest speed, while

the least steep part corresponds to the lowest speed.

19P

The distance D that the runner travels is represented by the area between the v vs t curve

and the time axis. Thus

D =

��1

2

�(2)(8) + (10� 2)(8) +

�1

2

�(12� 10)(8 + 4) + (16� 12)(4)

�m/s = 100m :

20E

(a) If v1 is the velocity at the beginning of a time interval (at time t1) and v2 is the velocity at

the end (at t2) then the average acceleration in the interval is given by �a = (v2�v1)=(t2�t1).Take t1 = 0, v1 = 18m/s, t2 = 2:4 s, and v2 = �30m/s. Then

�a =(�30� 18)m/s

(2:4� 0) s= �20m/s

2:

The minus sign indicates that the acceleration is opposite to the original direction of travel.

(b)

1 2 3 4 5

a

0

v

t

v

t


21E

22E

(a) Use v = dx=dt and a = d2x=dt2. For the interval AB, v is positive and a = 0. Similarly,

for BC: v is positive and a is negative; for CD: v =0 and a = 0; and for DE: v is negative

and a is positive.

(b) No.(c) No, becuase neither v nor a will be a�ected by shifting the origin of the x axis.

23E

See the previous problem. For the interval AB, v is positive and a is negative; for BC,

v = 0 and a = 0; for CD, v is positive and a is positive; and for DE, v is positive and

a = 0.

(b) No.(c) No, becuase neither v nor a will be a�ected by shifting the origin of the x axis.

24E

1

1

1

x

t (s ) t (s)

x

(a)

t (s ) t (s)

x

x

1

(b)

(c) (d)


25E

(e) For the speed to increase, the velocity and the acceleration must be in the same direction,

which is the case for situations (a), (b) and (d) above.

x (m)

100

400

2.0

t (s)1.0

200

300

3.0 4.0

0(a)

(b)

v (m/s)

50

200

2.0

t (s)1.0

100

150

3.0 4.00


26E

(a) No. The former is the square of the �rst derivative of x with respect to t, while the

latter is the second derivative of x with respect to t.

(b) The SI unit for the former is m2=s2, while it is m/s2 for the latter.

27E

(a) The average velocity during the �rst 3 s is given by �v = [x(3 s)�x(0)]=�t. Evaluate theexpression x = 50t+10t2 using t = 0 and t = 3 s: x(0) = 0 and x(3 s) = (50)(3)+(10)(3)2 =

240m. Thus

�v =(240� 0)m

3 s= 80m/s :

(b) The instantaneous velocity at time t is given by v = dx=dt = 50 + 20t, in m/s. At

t = 3:0 s, v = 50 + (20)(3:0) = 110m/s.

(c) The instantaneous acceleration at time t is given by a = dv=dt = 20m/s2. It is

constant, so the acceleration at any time is 20m/s2.

(d) and (e) The graph on the left below shows the coordinate x as a function of time (solid

curve). The dotted line marked (a) runs from t = 0, x = 0 to t = 3:0 s, x = 240m. Its

slope is the average velocity during the �rst 3 s of motion. The dotted line marked (b) istangent to x vs. t at t = 3:0 s. Its slope is the instantaneous velocity at t = 3:0 s. The

�gure on the right shows the instantaneous velocity as a function of time. Its slope is the

acceleration.

x (m)

0

-20

-401.0 2.5

t (s)-0.5

20

x (m)

20

-20

-80

1.0 2.5t (s)

-0.5

-40

-60

0

x (m)

-20

-80

1.0 2.5t (s)

-0.5

-40

-60

0


28E

(a) Let v(t) = dx=dt = 20� 15t2 = 0. Then t = 1:2 s.

(b) Let a(t) = dv=dt = �30t = 0. Then t = 0.

(c) At t > 0, the acceleration is always negative.

(d)

2 3 5 100

x (m) v (m/s)

0t (min)

2.2

2 3 5 8 1098 9

660

αβ

t (min)

γδ


29P

(a) During the interval from t = 2:00min to t = 8:00min,

�v =�x

�t=

(2:2m/s)(8:00min� 5:00min)

(8:00min� 2:00min)= 1:10m/s

and

�a =�v

�t=

2:20m/s

(8:00min� 2:00min)= 6:11mm/s2 :

(b) During the interval from t = 3:00min to t = 9:00min,

�v =�x

�t=

(2:2m/s)(9:00min� 5:00min)

(9:00min� 3:00min)= 1:47m/s

and

�a =�v

�t=

2:20m/s

(9:00min� 3:00min)= 6:11mm/s2 :

(c) The slopes of the lines labeled � through � in the two �gures below give the velocities

and the accelerations previously calculated. In particular, the slope of � is the average

velocity between t = 2:00min and 8:00min, the slope of � is the average velocity between

t = 3:00min and 9:00min; the slope of is the average acceleration between t = 2:00min

and 8:00min, and the slope of � is the average acceleration between t = 3:00min and

9:00min.

30P

(a) The instantaneous velocity and acceleration are given by v(t) = dx=dt = 6:0t2, and

a(t) = dv=dt = 12t, respectively, where v is in m/s and a is in m/s2. Between t = 1:0 s and

t = 2:0 s,

�v =�x

�t=

(2:0)[(2:0)3 � (1:0)3]m

(2:0� 1:0) s= 14m/s

x (m)

t (s)0

1 1.5 2 2.5

5

10

15

20

v1

v2

v (m/s)

t (s)

0

1 1.5 2 2.5

10

a1

a2

0.5-5

0.5-10

15a

v


and

�a =�v

�t=

(3)(2:0)[(2:0)2 � (1:0)2]m

(2:0� 1:0)m= 18m/s2 :

(b)

v(t = 1:0 s) = 6:0(1:0)2 = 6:0m/s ;

a(t = 1:0 s) = 12(1:0) = 12m/s2 ;

v(t = 2:0 s) = 6:0(2:0)2 = 24m/s ;

a(t = 2:0 s) = 12(2:0) = 24m/s2 :

(c) We see that v(t = 1:0 s) < �v < v(t = 2:0 s). This is because both the acceleration

and the velocity are in the same direcrtion (so the speed is increasing with time). Also

a(t = 1:0 s) < �a < a(t = 2:0 s), which is consistent with the fact that a(t) = 12:0t, which

increases with t.

(d) The various answers obtained in (a) through (c) are represented by the slopes in the

�gures below.


31P

(a) The velocity of the spot is given by

v(t) =dx

dt=

d

dt(9:00t� 0:750t3) = 9:00� 2:25t2:

Set v(t) = 0 in the equation above to obtain t = 2:00 s.

(b) x(t = 2 s) = (9:00)(2)� 0:750(2)3 = 12:0 cm:

(c)

a(t) = dv=dt =d

dt(9:00� 2:25t2) = �4:50t :

At t = 2:00 s, a(t) = �9:00m/s2.

(d) and (e) Since v(t = 2 s) = 0 and a(t = 2 s) < 0, it must be moving in the positive x

direction just before it stops, and in the negative x direction just after that.

Since v(t) > 0 for t > 2:00 s, we know that xmax = x(t = 2:00 s) = 12:0 cm: Thus the

spot never really reaches x = 15:0 cm. The next time it reaches an edge of the screen, its

coordinate is zero. Thus we let x(t) = 9:00t� 0:75t3 = 0 and get t = 3:46 s.

32P

(a) Since the units of et2 are those of length and the units of t are those of time, so the

units of e must be those of (length)/(time)2, or ft/s2. Since bt3 has units of length, b must

have units of (length)/(time)3, or ft/s3.

(b) When the particle reaches its maximum (or its minimum) coordinate its velocity is

zero. Since the velocity is given by v = dx=dt = 2et� 3bt2, v = 0 occurs for t = 0 and for

t = 2e=3b = (2)(3:0)=[3(1:0)] = 2:0 s. For t = 0, x = 0 and for t = 2:0 s; x = 4:0 ft. Reject

the �rst solution and accept the second.

(c) In the �rst 4 s the particle moves from the origin to x = 4:0 ft, turns around, and

goes back to x(4 s) = (3:0)(4)2 � (1:0)(4)3 = �16 ft. The total path length it travels is

4:0 + 4:0 + 16 = 24 ft.

(d) Its displacement is given by �x = x2 � x1, where x1 = 0 and x2 = �16 ft. Thus

�x = �16 ft.(e) The velocity is given by v = 6:0t� 3:0t2. Thus

v(1 s) = (6:0)(1:0)� (3:0)(1:0)2 = 3:0 ft/s

v(2 s) = (6:0)(2:0)� (3:0)(2:0)2 = 0:0

v(3 s) = (6:0)(3:0)� (3:0)(3:0)2 = �9:0 ft/sv(4 s) = (6:0)(4:0)� (3:0)(4:0)2 = �24 ft/s :

(f) The acceleration is given by a = dv=dt = 6:0� 6:0t. Thus

a(1 s) = 6:0� (6:0)(1:0) = 0:0

a(2 s) = 6:0� (6:0)(2:0) = �6:0 ft/s2

a(3 s) = 6:0� (6:0)(3:0) = �12 ft/s2

a(4 s) = 6:0� (6:0)(4:0) = �18 ft/s2 :


33E

The time it takes is

�t =�v

a=

100 km/h

50m/s2=

(100 km/h)(1 h=3600 s)(1000m/km)

50m/s2= 0:556 s :

(Question: How does this compare with your Ferrari?)

34E

Use v = v0+at, an equation that is valid for motion with constant acceleration. Take t = 0

to be the time when the velocity is +9:6m/s. Then v0 = 9:6m/s.

(a) Since we wish to calculate the velocity for a time before t = 0, the value we use for t is

negative: t = �2:5 s. Thus

v = (9:6m/s) + (3:2m/s2)(�2:5 s) = 1:6m/s :

(b) Now t = +2:5 s and

v = (9:6m/s) + (3:2m/s2)(2:5 s) = 18m/s :

35E

For the automobile

a =�v

�t=

(55� 25)( km/h)(1 h=3600 s)(1000m/km)

(0:50min)(60 s/min)= 0:28m/s2 ;

and for the bicycle

a =�v

�t=

(30� 0)( km/h)(1 h=3600 s)(1000m/km)

(0:50min)(60 s/min)= 0:28m/s2 :

36E

Solve v = v0+at for t: t = (v�v0)=a. Substitute v = (0:1)(3:0�108m/s) = 3:0�107m/s,

v0 = 0, and a = 9:8m/s2. The result is 3:1� 106 s.

(b) Evaluate x = x0+v0t+12at2, with x0 = 0. The result is x = 1

2(9:8m/s

2)(3:1�106 s)2 =

4:6� 1013m.

x (cm)

2.5

10

20

t (ns)10

5.0

7.5

30 400

v (10 6 m/s)

2.0

8.0

20

t (ns)10

4.0

6.0

30 400

28

CHAPTER 2 MOTION ALONG A STRAIGHT LINE

37EUse v 2� v 2

0 = 2a�x: to solve for a. Its minimum value isamin = v 2� v 2

02�xmax= (360 km/h)22(1:80 km) = 2:8m/s2 :

38E(a) The velocity is given by v = v0 + at. The time to stop can be found by setting v = 0

and solving for t: t = �v0=a. Substitute this expression into x = v0t + 12 at2 to obtain

x = � 12 v 20=a. [Alternatively, you might use v 2 = v 2

0 +2a(x� x0).] Use v0 = 5:00� 106m/s

and a = �1:25 � 1014m/s 2. Notice that since the muon slows the initial velocity and the

acceleration must have opposite signs. The result isx = � 1

2(5:00� 106m/s)2(�1:25� 1014m/s 2)

= 0:100m :

(b) Here are the graphs of the coordinate x and velocity v of the muon from the time it

enters the �eld to the time it stops:

39ESolve v 2 = v 2

0 + 2a(x � x0) for a. Take x0 = 0. Then a = (v 2 � v 20 )=2x. Use v0 =

1:50� 105m/s, v = 5:70� 106m/s, and x = 1:0 cm = 0:10m. The result is

a = (5:70� 106m/s)2� (1:50� 105m/s)22(0:010m)

= 1:62� 1015m/s 2 :

61.5

5

x (m) v (m/s)

t (s)t (s)

24.6

5


40E

(a) Using v2 � v20 = 2a�x, and noting that v = 0, we �nd

jaj = v202�x

=(60mi/h)2[(1609m/mi)=(3600 s/h)]2

(2)(43)m

= 8:3m/s2 =

�8:3m/s2

9:8m/s2

�g = 0:85g :

(b) The stopping time is t = v0=a = 3:2 s; which corresponds to t=T = 3:2 s=400ms = 8

\reaction times".

41E

The acceleration is

a =�v

�t=

(102 km/h)(1000m/km)(1 h=3600 s)

1:4 s= 202:4m/s2 :

In terms of g, this is

a =

�202:4m/s2

9:8m/s2

�g = 21g :

42E

(a) The time it takes to stop is

t =v0

a=

24:6m/s

4:92m/s2= 5:00 s :

(b) The distance traveled is

�x =v202a

=(24:6m/s)2

(2)(4:92m/s2)= 61:5m/s2 :

(c)

x (ft)

300

2.0t (s)

1.0

100

200

3.00


43E

The acceleration of the sled is

a =�v

�t=

(1600 km/h)(1000m/km)(1 h=3600 s)

1:8 s= 2:5� 102m/s2 :

In terms of g, this is

a =

�2:5� 102m/s2

9:8m/s2

�g = 25g :

(b) The distance traveled is s = 12at2 = (2:5� 102m/s2)(1:8 s)2=2 = 4:0� 102m:

44E

(a) Assume the acceleration is uniform and solve v = v0 + at for t: t(v� v0)=a. Substitute

v0 = 85mi/h = 125 ft/s, v = 55mi/h = 80:7 ft/s, and a = �17 ft/s2. Note that since the

car is slowing down the initial velocity and the acceleration must have opposite signs. The

result is

t =(80:7� 125) ft/s

(�17 ft/s2) = 2:6 s :

(b) Suppose the coordinate of the car s

x = 0 when the brakes are applied (at

time t = 0). Then the coordinate of the

car as a function of time is given by x =

(125 ft/s)t � 12(17 ft/s2)t2. This function

is plotted from t = 0 to t = 2:6 s on the

graph to the right.

45E

First, calculate the deceleration a of the motorcycle: a = �v=�t = (15m/s� 30m/s)=3:0 s

= �5:0m/s2: The distance �x traveled by the motorcycle before it stops can then be

obtained from v2 = v20 + 2a�x:

�x =v2 � v202a

=02 � (30m/s)2

2(�5:0m/s2)= 90m :

46P

(a) �a =�v

�t=

(60 km/h)(1000m/km)(1 h=3600 s)

5:4 s= 3:1m/s2:

(b) x =1

2�at2 =

(3:1m/s2)(5:4 s)2

2= 45m :

x (m ) v (m/s)

t (s)t (s)

250

90

0 6 10

30

60

6 10


(c)

t =

r2x

�a=

s2(250m)

3:1m/s2= 13 s :

47P

(a) From v22 � v21 = 2a�x, we �nd the acceleration to be

a =v22 � v212�x

=(502 � 302)m2= s2

(2)(160)m= 5:0m/s :

(b) The time required is

�t =�v

a=

(50� 30)m/s

5m/s2= 4:0 s :

(c) The time it takes for v to reach 30m/s is

t =v

a=

30m/s

5:0m/s= 6:0 s :

(d) The distance moved is

x =1

2at2 =

1

2(5:0m/s2)(6:0 s)2 = 90m :

(e)

v (m/s)

15

10t (s)

5

5

10

-5 0

x (m)

60

10t (s)

5

20

40

-5


48P

(a) Take x0 = 0, and solve x = v0t+12at2 for a: a = 2(x� v0t)=t

2. Substitute x = 24:0m,

v0 = 56:0 km/h = 15:55m/s, and t = 2:00 s. The result is

a =2 [24:0m� (15:55m/s)(2:00 s)]

(2:00 s)2= �3:56m/s

2:

The minus sign indicates that the acceleration is opposite the direction of motion of the

car. The car is slowing down.

(b) Evaluate v = v0 + at. You should get v = 15:55m/s� (3:56m/s2)(2:00 s) = 8:43m/s =

30:3 km/h.

49P

(a) Let x = 0 at the �rst point and take the time to be 0 when the car is there. Use

v = v0+ at to eliminate a from x = v0t+12at2. The �rst equation yields a = (v� v0)=t, so

x = v0t +12(v � v0)t =

12(v + v0)t. Solve for v0: v0 = (2x � vt)=t. Substitute x = 60:0m,

v = 15:0m/s, and t = 6:00 s. Your result should be

v0 =2(60:0m)� (15:0m/s)(6:00 s)

6:00 s= 5:00m/s :

(b) Substitute v = 15:0m/s, v0 = 5:00m/s, and t = 6:00 s into a = (v � v0)=t. The result

is a = (15:0m/s� 5:00m/s)=(6:00 s) = 1:67m/s2.

(c) Substitute v = 0, v0 = 5:0m/s, and a = 1:67m/s2into v2 = v20 + 2ax. Solve for x:

x = � v202a

= � (5:0m/s)2

2(1:67m/s2)= �7:50m :

(d) To draw the graphs you need to know the time at which the car is at rest. Solve

v = v0 + at = 0 for t: t = �v0=a = �(5:00m/s)=(1:67m/s2) = �3:0 s. Your graphs should

look like this:

1100

550

0 30.3 60.6 0 30.3 60.6

36.3

x (m)

t (s) t (s)

v (m/s)

-1.2

0

1.2

30.3 60.6

a (m/s2)

t (s )


50P

(a) The time t1 it takes to cover the �rst half of the distance is given by at2=2 = s=2, which

yields

t1 =

rs

a=

s1100m

1:2m/s2= 30:3 s :

Since it takes an equal amount of time to slow down to zero speed, the total time of travel

is T = 2t1 = 60:6 s.

(b) vmax = at1 = (1:2m/s2)(30:3 s) = 36:3m/s:

(c)


51P

Let tr be the reaction time and tb be the braking time. Then the total distance moved

by the car is given by x = v0tr + v0tb +12at2b , where v0 is the initial velocity and a is the

acceleration. After the brakes are applied the velocity of the car is given by v = v0 + atb.

Use this equation, with v = 0, to eliminate tb from the �rst equation. According to the

second equation tb = �v0=a, so x = v0tr � v20=a+12v20=a = v0tr � 1

2v20=a.

Write this equation twice, once for each of the two di�erent initial velocities: x1 = v01tr �12v201=a and x2 = v02tr � 1

2v202=a. Solve these equations simultaneously for tr and a. You

should get

tr =v202x1 � v201x2

v01v02(v02 � v01)

and

a = �1

2

v02v201 � v01v

202

v02x1 � v01x2:

Substitute x1 = 186 ft, v01 = 50mi/h = 73:4 ft/s, x2 = 80 ft, and v02 = 30mi/h = 44:0 ft/s.

The results are

tr =(44:0 ft/s)2(186 ft)� (73:4 ft/s)2(80 ft)

(73:4 ft/s)(44:0 ft/s)(73:4 ft/s� 44:0 ft/s)= 0:74 s

and

a = �1

2

(44:0 ft/s)(73:4 ft/s)2 � (73:4 ft/s)(44:0 ft/s)2

(44:0 ft/s)(186 ft)� (73:4 ft/s)(80 ft)= �20 ft/s2 :

52P

(a) The initial speed of the car, in m/s, is v0 = (35mi/h)(1609m/mi)(1 h=3600 s) =

15:6m/s. Denote L = 40m and T = 2:8 s. If you decide to brake to a stop, then by

the time you start applying the brake the car has already traveled further towards the

intersection by as much as d = v0tr, where tr is your reaction time. Thus at the moment

you start to apply the brake the distance between the car and the intersection is

D = L� d = 40m� (15:6m/s)(0:75 s) = 28m :

At this point, the car starts to decelerate. Suppose it will travel further by a distance

�x before coming to a complete stop, then v20 = 2a�x, where a is the magnitude of the

deceleration. Solve for �x:

�x =v202a

=(15:6m/s)2

2(17 ft/s2)(0:3048m/ft)= 24m :

Since D � �x = 28m � 24m = 4m > 0, the car will come to a stop before entering the

intersection.


If you choose to continue to move, then you will be able to reach the intersection in

t = L=v0 = 40m=(15:6m/s) = 2:6 s. Since the yellow light will last for T = 2:8 s, you can

make it into the intersection while the light is still yellow. Thus either strategy will work.

(b) In this case you can follow the same lines of argument as part (a) above, with the new

values for L and T being L = 32m and T = 1:8 s. As a result, you will �nd that, if you

choose to apply the brake, you wouldn't be able to stop your car untill its front has already

traveled 4m into the intersection. And if you choose to continue moving, the tra�c light

would have already turned yellow while your car is still about 4m short of reaching the

intersection. So neither strategy will work in this case!

53P

(a) Let the reaction time of the driver be tr. The reaction distance is then Dr = v0tr,

where v0 is the initial speed of the car . The braking distance Db satis�es

v20 = 2jajDb ;

where a is the deceleration of the car. From the above equations, we �nd, for v0 = 10m/s,

tr = Dr=v0 = 7:5m=10m/s = 0:75 s and

jaj = v202Db

=(10m/s)2

2(5:0m)= 10m/s2 :

You can easily check that tr and jaj are the same for v0 = 20m/s and v0 = 30m/s.

(b) For v0 = 25m/s, the stopping distance Ds is

Ds = Dr +Db = v0tr +v202jaj

= (25m/s)(0:75 s) +(25m/s)2

2(10m/s2)

= 50m :

54P

(a) Since the maximum acceleration and deceleration have the same magnitude, from sym-

metry consideration we know that the maximum speed is attained when the train is mide-

way between the two stations, which are separated by a distance D. Thus v2max � v20 =

v2max = 2a�x = 2a(D=2), which gives

vmax =pDa =

p(806m)(1:34m/s)2 = 32:9m/s :

(b) The time it takes for the train to accelerate to vmax is given by

t =vmax

a=

32:9m/s

1:34m/s2= 24:5 s :

t (s)t (s )

v (m/s)x (m )

806

24.5 49 690 0

32.9

24.5 49 69

v=0

1.34

-1.34

24.5 49 69t (s )

a(m/s2)

0

a=0

stop

36


Since it takes an equal amount of time to accelerate and decelerate, the total travel time

between the two stations is 2t = 49:0 s.

(c) The maximum average speed is

�vmax = D

t+ 20 s= 806m(49 + 20) s

= 11:7m/s :

(d)


55P

(a) Use v2max = 2a�x to solve for �x:

�x =v2max

2a=

(1000 ft=60 s)2

(2)(4 ft/s2)= 34:7 ft :

(b) The time it takes to accelerate and decelerate are both given by �t1 = vmax=a, while

it remains at the top speed for �t2 = (D� 2�x)=vmax, where �x is given in (a). Thus thetotal time it takes is

t = 2�t1 +�t2 =2vmax

a+D � 2�x

vmax

=D

vmax

+vmax

a

=624 ft

1000 ft=60 s+

1000 ft=60 s

4:0 ft/s2= 41:6 s :

56P

(a) Let the required time be t. By this time, the distances the two vehicles have traveled

must be the same. Let the acceleration of the automobile be a and the speed of the truck

be v. Then

s =1

2at2 = vt ;

which gives

t =2v

a=

2(9:5m/s)

2:2m/s2= 8:6 s :

Thus

s = vt = (9:5m/s)(8:6 s) = 82m :

(b) The speed of the car is

vc = at = (2:2m/s2)(8:6 s) = 19m/s :

57P

(a) From the point of view of the locomotive (refered to as 1), the train will be closing in on

it as long as the speed of the train (refered to as 2) is greater than that of the locomotive. So

the closest distance between the two will be reached the moment when they have identical

speeds. Let the time of this instant be t. Then v1 = v2 � a2t, or

t =v2 � v1

a2;

x (mi)

0.42

t

train (accident avoided)

train (collision)

locomotive


where v2 is the initial speed of the train. To ensure that no collision takes place, the

separation between the two must not be negative at time t, i.e., v1t�(v2t� 12a2t

2)+0:42mi �0; or

v1(v2 � v1)

a2� v2(v2 � v1)

a2+

(v2 � v1)2

2a2+ 0:42mi

= � (v2 � v1)2

2a2+ 0:42mi � 0 ;

which gives

a2 � (v2 � v1)2

2(0:42mi)=

(100mi/h� 18mi/h)2

2(0:42mi/h)= 3:26 ft/s2 :

(b)

58P

The stopping distance for the �rst train is given by

�x1 =v212a

=[(72 km/h)(103m/km)(1 h=3600 s)]2

(2)(1:0)m/s2= 200m :

Similarly, for the second train

�x2 =v222a

=[(144 km/h)(103m/km)(1 h=3600 s)]2

(2)(1:0)m/s2= 800m :

Since �x1 +�x2 = 1000m > 950m, there will indeed be a collision.

v

t

y (m)

02

t (s)1

-20

-10

3

-30

v (m/s)

02

t (s)1

-20

-10

3

-30

a(m/s2)

02

t (s)1

-10

-5

3


59P

60E

(a) Take the y axis to be positive in the upward direction and take t = 0 and y = 0

at the point from which the wrench was dropped. If h is the height from which it was

dropped and the y axis is upward then the ground is at y = �h. Solve v2 = v20 + 2gh for

h: h = (v2 � v20)=2g. Substitute v0 = 0, v = �24m/s, and g = 9:8m/s2:

h =(24m/s)2

2(9:8m/s2)= 29:4m :

(b) Solve v = v0 � gt for t: t = (v0 � v)=g = (24m/s)=(9:8m/s2) = 2:45 s.

(c)

y (m)

60

4t (s)

2

20

40

v (m/s)

40

4t (s)

20

20

6

-20

a(m/s2)

04

t (s)2

-10

-5

6

06 8

-40

8

8


61E

(a) At the highest point the velocity of the ball is instantaneously zero. Take the y axis to

be upward, set v = 0 in v2 = v20�2gy, and solve for v0: v0 =p2gy. Substitute g = 9:8m/s

2

and y = 50m to get v0 =

q2(9:8m/s

2)(50m) = 31m/s.

(b) It will be in the air until y = 0 again. Solve y = v0t� 12gt2 for t. Since y = 0 the two

solutions are t = 0 and t = 2v0=g. Reject the �rst and accept the second: t = 2v0=g =

2(31m/s)=(9:8m/s2) = 6:4 s.

(c)

62E

From v2 = 2a�x = 2gh, we �nd

v =p2gh =

p(2)(9:80m/s2)(1700m) = 183m/s :

This is a much higher speed than that of actual raindrops. It's hard to say whether it can

kill, but if I were you, I'd de�nitely stay under a sturdy roof in a rainstorm like this!


63E

(a) Use v2 = v20 + 2gy to solve for v:

v =p2gy =

p2(9:8m/s2)(120m) = 48:5m/s :

(b) The time of ight t satis�es y = gt2=2, so

t =

r2y

g=

s2(120m)

9:8m/s2= 4:95 s :

(c) The speed is

v0 =

s2g

�y

2

�=pgy =

p(9:8m/s2)(120m) = 34:3m/s :

(d) Use v0 = gt0. The time t0 is then

t0 =v0

g=

34:3m/s2

9:8m/s2= 3:50 s :

64E

The time t that it takes for the stone to reach the ground satis�es

y = 30:0m = v0t+1

2gt2 = (12:0m/s)t+

1

2(9:80m/s2)t2 ;

which gives t = 1:54 s.

(b) The speed at impact is given by

v = v0 + gt = 12:0m/s + (9:80m/s2)(1:54 s)2 = 27:1m/s :

65E

(a)

t =

r2y

g=

s2(145m)

9:8m/s2= 5:44 s :

(b) v = gt = (9:8m/s2)(5:44 s) = 53:3m/s:

(c) The distance �y satis�es v2 = 2a�y = 2(25g)�y, which gives

�y =v2

50g=

(53:3m/s)2

50(9:8m/s2)= 5:80m :

y v a

tt

t

fuel exhausted hits ground fuel exhausted fuel exhausted hits ground

ymax

42


66E

Assume that the rocket is being

propelled by a constant force F , in which case the acce

l-

eration a of the rocketincreases as m

ore and more fuel get burnt, since th

e mass m of the

rocket is decreasing (Note that a

= F=m). The �gures are shown

below.

67E

Take the y axis to be upward, set v0 = 0, and solve y = v0t� 1

2gt2 for t: t =

p�2y=g.

(a) For this part y = �50m, so

t =

s�2(�

50m)

9:8m/s2= 3:2 s :

(b) For this part set y = �100m, so

t =

s�2(�

100m)

9:8m/s2

= 4:5 s :

The di�erence is the time taken to fall the secon

d 50m: 4:5 s� 3:2 s = 1:3 s.

68P

(a) Take they axis to be positive u

pward and use y = v0t � 12gt2, with y = 0:544m and

t = 0:200 s. Solvefor v0:

v0 =y +

12gt2

t=

(0:544m) +12(9:8m/s

2)(0:200 s)2

0:200 s

= 3:70m/s :

(b) Use v = v0 � gt = (3:70m/s)� (9:8m/s2)(0:200 s) = 1:74m/s.


(c) Use v2 = v20 � 2gy, with v = 0. Solve for y:

y =v202g

=(3:7m/s)2

2(9:8m/s2)= 0:698m :

It goes (0:698m� 0:544m) = 0:154m higher.

69P

Assume that it takes time t for the object to fall. Then from h = gt2=2 we �nd

t =

s2h

g=

s2(45m)

9:8m/s2= 3:0 s :

In the mean time, the boat has moved a distance D = 12m. Thus the speed of the boat is

vb = D=t = (12m)=(3:0 s) = 4:0m/s:

70P

(a) At the point where its fuel gets exhausted, the rocket has reached a height of

y0 =1

2at2 =

(4:00m/s2)(6:00 s)2

2= 72:0m :

The speed of the rocket at this instant is

v0 = at = (4:00m/s2)(6:00 s) = 24:0m/s :

The additional height y1 the rocket can attain on top of y0 is then given by v20 = 2gy1,

which gives

y1 =v202g

=(24:0m/s)2

2(9:80m/s2)= 29:1m :

The total height the rocket attains is thus 72:0m + 29:1m = 101m.

(b) The time of ight after the fuel gets exhausted is given by

�y0 = v0t� 1

2gt2 ;

or

�72:0m = (24:0m/s)t� 1

2(9:80m/s2)t2 :

Solve for t to obtain t = 7:00 s. The total time of ight is therefore 7:00 s + 6:00 s = 13:0 s.


71P

(a) Let the height reached by the player be y. The initial speed v0 of the player is

v0 =p2gy =

p2(9:80m/s2)(0:760m) = 3:86m/s:

As the player reaches y1 = 76:0 cm�15:0 cm = 61:0 cm, his speed v1 satis�es v20�v21 = 2gy1,

which gives

v1 =

qv20 � 2gy1 =

p(3:86m/s)2 � 2(9:80m/s2)(0:610m) = 1:72m/s :

The time t1 that the player spends in the top 15:0 cm of this jump then satis�es v1�gt1=2 =0, or

t1 =2v1

g=

2(1:72m/s)

9:80m/s2= 0:350 s = 350ms ;

where the factor of 2 in the expression for t1 accounts for the time for him to fall.

(b) The time t2 for the player to jump to a height of 15:0 cm satis�es

15:0 cm = v0t2 � 1

2gt22 = (3:86m/s)t2 � (9:80m/s2)t22

2;

which yields t2 = 0:0820 s = 82:0ms.

72P

Let the speeds of the ball at levels L and U be vL and vU , respectively. Then since it takes

�TU=2 to go from hU to the maximum height hmax,

hmax � hU = vU

��TU

2

��1

2

��TU

2

�2

:

Similarly,

hmax � hL = vL

��TL

2

��1

2

��TL

2

�2

:

Taking the di�erence of the above two equations, we get

H = hU � hL =

�vL�TL

2� vU�TU

2

�� g

��TL

2 ��TU2

8

�:

Now substitute vU = g(�TU=2) and vL = g(�TL=2) into the expression for H above to

obtain H = g(�T 2L ��T 2

U )=8, or

g =8H

�T 2L ��T 2

U

:


73P

The speed of the ball just before hitting the ground is given by v =p2gy, where y is the

initial height from which the ball is dropped. The average acceleration is then

�a =�v

�t=

p2gy

�t=

p2(9:80m/s2)(15:0m)

20:0ms= 857m/s2 :

The direction of the acceleration is upward.

74P

(a) For the ball which starts with a speed v0 and reaches a speed of v just before hitting

the ground, we may write v2 � v20 = 2gH, where H is the distance it traveled. Thus

v =pv20 + 2gh: The direction of the velocity is downward.

(b) The time it takes is given by

t =v � v0

g=

pv20 + 2gh� v0

g:

(c) If the ball is thrown upward, the answer to (a) will remain the same, while the answer

to (b) will be greater, with the new time-of- ight t0 being

t0 = t+2v0

g=

pv20 + 2gh+ v0

g:

75P

(a) The distance D from the lower dot to the mark corresponding to a certain reaction

time t is given by D = gt2=2. Thus for t1 = 50:0ms

D1 =(9:8m/s2)(50:0ms)2

2= 1:23 cm :

(b) For t2 = 100ms

D2 =(9:8m/s2)(100ms)2

2= 4D1 ;

for t3 = 150ms

D3 =(9:8m/s2)(150ms)2

2= 9D1 ;

for t4 = 200ms

D4 =(9:8m/s2)(200ms)2

2= 16D1 ;

and for t4 = 250ms

D5 =(9:8m/s2)(250ms)2

2= 25D1 :


76P

For the �rst throw H1 = gt21=2, while for the second throw H2 = gt22=2 = g(2t1)2=2 = 4H1.

Thus one has to throw four times as high to make the ball stay in the air twice as long.

77P

(a) Use v2 = v20 + 2a(x� x0): Thus�v

2

�2

= v2 + 2(�g)(x� x0) = v2 + 2(�9:80m/s2)(3:00m) ;

which gives v = 8:85m/s.

(b) The height reached above B is given by

y =(v=2)2

2g=

(8:85m/s)2

8(9:80m/s2)= 1:00m :

78P

The average acceleration during the time when ball is in contact with the oor is given

by �a = (v2 � v1)=�t, where v1 is its velocity just before striking the oor and v2 is its

velocity just after leaving the oor. Take the y axis to upward and place the origin at

the point where the ball is dropped. To �nd the velocity just before hitting the oor, use

v21 = v20 � 2gy, where v0 = 0 and y = �4:00m. The result is

v1 =p�2gy =

q�2(9:8m/s

2)(�4:00m) = �8:85m/s :

The negative square root is used because the ball is traveling downward. To �nd the

velocity just after hitting the oor, use v2 = v22 � 2g(y� y0), with v = 0, y = �1:00m, and

y0 = �4:00m. The result is

v2 =p2g(y � y0) =

q2(9:8m/s

2)(�1:00m + 4:00m) = 7:67m/s :

The average acceleration is

�a =v2 � v1

�t=

7:67m/s + 8:85m/s

10:0� 10�3 s= 1:65� 103m/s

2:

y3

y2

y1

faucet

y


79P

Let the time of ight of the nth drop be tn (n = 2; 3; 4), then

t1 = 3t3 and t2 = 2t3. Since yn = 12gt2n, we have

y3

y1=

t23t21

=

�1

3

�2

=1

9;

which gives y3 = y1=9 = 22 cm. Also,

y2

y1=

t22t21

=

�2

3

�2

=4

9;

which gives y2 = 4y1=9 = 89 cm.

80P

(a) Let the height of the diving board be h, the depth of the lake be D, and the total time

for the ball to descend be T . The speed of the ball as it reaches the surface of the lake

is then v =p2gh. The time the ball spends in the air is t1 =

p2h=g, while the time it

spends descending in the lake is t2 = D=v = D=p2gh. Thus

T = t1 + t2 =

s2h

g+

Dp2gh

;

which gives

D =p2ghT � 2h =

p(2)(9:80m/s2)(5:20m)(4:80 s)� (2)(5:20m) = 38:1m :

(b) The average speed is

�v =D + h

T=

38:1m + 5:20m

4:80 s= 9:02m/s :

(c) In this case, the ball has to be thrown upward. Following the solution to part (c) in

74P, we have gT =p2g + v20 + v0, which yields

v0 =gT

2� h+D

T=

(9:80m/s2)(4:80 s)

2� 38:6m + 5:20m

4:80 s= 14:5m/s :

81P

Let h be the height of the fall and t be the time of the fall. Take the y axis to be upward

and place the origin at the point of release. Then �h = �12gt2: Since it falls half the

distance down in the last second, it falls the �rst half in time t� 1, where t is in seconds.

Thus �h=2 = �12g(t � 1)2. Solve these two equations simultaneously for t and h. First

2.00 3.001.00

144

0t (s)

y (ft)

2nd stone

1st stone


substitute 12gt2 for h in h = g(t � 1)2 to obtain t2 = 2(t � 1)2. A little manipulation

gives the quadratic equation t2 � 4t + 2 = 0. It has the solutions t = 2 � p2 = +3:41 s

and �0:586 s. You want the positive solution. To �nd the height substitute this value for

t into h = 12gt2 to obtain h = 1

2(9:8m/s

2)(3:41 s)2 = 57m. The negative solution for t

corresponds to a ball thrown upward, during the time before it reaches its highest point.

82P

The speed of the woman just before she lands in the ventilator box is given by v =p2gh =p

(2)(32 ft/s2)(144 ft) = 96 ft/s. The acceleration she experiences as she crashes onto the

box is then

a =v2

2�x=

(96 ft/s)2

(2)(18 in.)= 3072 ft/s2 ;

which is equivalent to [(3072 ft/s2)=(32 ft/s2)]g = 96g.

83P

(a) The time of ight t1 for the �rst stone is given by

t1 =

r2y1

g=

s2(144 ft)

32:0 ft/s2= 3:00 s :

Thus the time of ight t2 for the second stone must be t2 = t1 � 1:00 s = 2:00 s. Let its

initial speed be v0, then

y1 = v0t2 +1

2gt22 ;

or

144 ft = v0(2:00 s) +

�32:0 ft/s2

2

�(2:00 s)2 ;

which gives v0 = 40:0 ft/s.

(b)


84P

(a) The time t1 during which the parachutist is in free fall is given by h1 = 50m = gt21=2 =

(9:80m/s2)t21=2, which yields t1 = 3:2 s. The speed of the parachutist just before he opens

the parachute is given by v21 = 2gh1, or

v1 =p2gh1 =

p(2)(9:80m/s2)(50m) = 31m/s :

If the �nal speed is v2, then the time interval t2 between the opening of the parachute and

the arrival of the parachutist at the ground level is

t2 =v1 � v2

a=

31m/s� 3:0m/s

2m/s2= 14 s :

The total time of ight T is therefore T = t1 + t2 = 17 s.

(b) The distance through which the parachutist falls after the parachute is opened is given

by

h =v21 � v222a

=(31m/s)2 � (3:0m/s)2

(2)(2:0m/s2)= 240m :

Thus the fall begins at a height of H = 50m + h = 290m.

85P

Take the y axis to be upward and place the origin at the point from which the objects are

dropped. Suppose the �rst object is dropped at time t = 0. Then its coordinate is given

by y1 = �12gt2 and, since the second object starts 1:0 s later, its coordinate is given by

y2 = �12g(t � 1)2, where t is in seconds. You want the time for which y2 � y1 = 10m.

Solve �12g(t� 1)2+ 1

2gt2 = 10, with g = 9:8m/s

2. After writing (t� 1)2 as t2� 2t+1 and

canceling the terms in t2 you should get t = (10=g) + 1 = 1:5 s.

86P

Let the vertical distances between Jim's and Clara's feet and the jump-o� level be HJ and

HC , respectively. At the instant this photo was taken, Clara has fallen for a time TC , while

Jim has fallen for TJ . Thus HJ = gT 2J=2 and HC = gT 2

C=2. Measuring directly from the

photo, we get HJ � 3:6m and HC � 6:3m, which yields TJ � 0:85 s and TJ � 1:1 s. Thus

Jim has waited for �T = TC � TJ � 0:3 s:

87P

(a) Take the y axis to be upward and place the origin on the ground, under the balloon.

Since the package is dropped, its initial velocity is the same as the velocity of the balloon,

+12m/s. The initial coordinate of the package is y0 = 80m; when it hits the ground its

coordinate is 0. Solve y = y0 + v0t� 12gt2 for t:

t =v0

g�sv20g2

+2y0

g=

12m/s

9:8m/s2+

s(12m/s)2

(9:8m/s2)2

+2(80m)

9:8m/s2= 5:4 s ;

where the positive solution was used. A negative value for t corresponds to a time before

the package was dropped.

0 (roof)

1 (top of window)

2 (bottom window)

3 (ground)

y


(b) Use v = v0 � gt = 12m/s� (9:8m/s2)(5:4 s) = �41m/s. Its speed is 41m/s.

88P

(a) Take the y axis to be upward and place the origin at the point where the ball is thrown.

Take the time to be 0 at the instant the ball is thrown. Then the velocity of the ball

is given by v = v0 � gt and at the highest point v = 0. Solve 0 = v0 � gt for t and

substitute the result into y = y0 + v0t � 12gt2 to obtain y = 1

2v20=g. The initial velocity is

20m/s + 10m/s = 30m/s. The result is

y =(30m/s)2

2(9:8m/s2)= 46m :

It is then 46m + 30m = 76m above the ground.

(b) The coordinate of the ball is given by y = v0t � 12gt2. When the ball is thrown the

oor of the elevator has coordinate yf0 = �2:0m and the elevator is moving upward with

constant velocity vf . Thus the coordinate of the oor at time t is given by yf = yf0 + vf t.

You want to solve for the time when the coordinate of the ball and elevator oor are the

same: y = yf . This means v0t� 12gt2 = yf0 + vf t. The solution to this quadratic equation

is

t =(v0 � vf )

g�s

(v0 � vf )2

g2� 2yf0

g

=(30m/s� 10m/s)

9:8m/s2

+

s(30m/s� 10m=s)2

(9:8m/s2)2

� 2(�2:0m)

(9:8m/s2)= 4:2 s ;

where the positive root was used. The negative root corresponds to a time before the ball

is thrown.

89P

From the graph, we may write

v2 = v1 + g(t2 � t1)

y2 � y1 = v1(t2 � t1) +1

2g(t2 � t1)

2 :

Substitute t2 � t1 = 0:125 s, y2 � y1 = 1:20m, and

g = 9:80m/s2 into the equations above and solve for

v1 and v2: v1 = 8:99m/s, v2 = 10:2m/s. Thus

y1 =v212g

= 4:12m

and

y2 = y1 + 1:20m = 5:32m :


Also

y3 � y2 = v2(t3 � t2) +1

2g(t3 � t2)

2 ;

where t3 � t2 = 2:00 s=2 = 1:00 s: Thus the height of the building is

y3 = 5:32m + (10:2m/s)(1:00 s) +(9:80m/s2)(1:00 s=2)

2= 20:4m :

90P

The time t the pot spends going up and going down past the window of length L is 0:25 s

apiece. Assume that the pot passed the bottom of the window with a speed v1 and it

passed the top of the window with a speed v2. Then

v1 � v2 = gt

and

�v =v1 + v2

2=

L

t:

Solve for v2: v2 = L=t� gt=2. The distance H the pot goes above the top of the window

is therefore

H =v222g

=(L=t� gt=2)2

2g=

[2:00m=0:25 s� (9:80m/s2)(0:25 s)=2]2

(2)(9:80m/s2)= 2:34m :

91

(a) d = v2i =2a0 + TRvi;

(b) 9:0m/s2;

(c) 0:66 s

92

(a) d = 12a0t2;

(c) 7:0m/s2

93

(a) v2j = 2a0d0(j � 1) + v21 ;

(c) 7:0m/s2;

(d) 14m

Various mathematics packages on a personal computer can be used to solve the problems in the Electronic Computation section. As an example this problem is solved using the "Symbolic Calculation" feature on Mathcad. Load the symbolic processor after you open up the Mathcad program. The following is an actual computer printout of a Mathcad document desinged to solve this problem. (a) To obtain an expression for the velocity v as a fuinction of time, write down the expression for x(t) first:

x( )t ( )32.0 ..24.0 t2 e.0.0300 t

Use Mathcad to evaluate symbolically the derivative of x(t) with respect to time, which is the velocity:

v(t) =d

dt( )32.0 ..24.0 t2 e

.0.0300 t

with the result v( )t ..48.0 t exp( )..3.0 10 2 t ...72 t2 exp( )..3.0 10 2 t

To obtain the acceleration as a function of time, take the derivative of v(t) with respect to time:

a(t) = ddt

( )..48.0 t exp( )..3.0 10 2 t ...72 t2 exp( )..3.0 10 2 t

which yields

a( )t .48.0 exp( )..3.0 10 2 t ..2.88 t exp( )..3.0 10 2 t ...2.16 10 2 t2 exp( )..3.0 10 2 t

(b ) Now plot x, v and a as functions of time t:

t . .,0 .1 100

0 20 40 60 80 1005000

0

5000

1 104

1.5 104

x( )t

t


94

t ..,0 .1 100

0 20 40 60 80 100200

100

0

100

200

300

400

v( )t

t

t ..,0 .1 100

0 20 40 60 80 10020

0

20

40

60

a( )t

t

(c) To find the time when x(t)=0, use the "root" feature in Mathcad. Let the time be t1when x reaches zero. To find t1, first give it an initial "guess" value, say t1 = 0.1 (as you can see clearly from the x vs t curve plotted above). Then the value of t1 can be found as follows:


t 1 1 =root ,32.0 ..24.0 t 12 e

.0.0300 t 1 t 1 1.175

So the time is t1 = 1.175 sec when x = 0. Once this is known, the velocity v and acceleration a at that moment are easily obtained:

=v( )1.175 53.487 =a( )1.175 43.099

(d ) Similar to part (c) above, we first find the time t2 when v = 0. From the graph of v vs t we give an intial guess of t2 = 65. Then

t 2 65 =root ,..48.0 t 2 exp ..3.0 10 2 t 2...72 t 2

2 exp ..3.0 10 2 t 2 t 2 66.667

The actual time is then 66.7 sec. The corresponding values for x and a are

=x( )66.667 1.44 104

=a( )66.667 6.496

(As another useful feature, Mathcad does allow the quantities to carry their units in the calculation. It can keep track and figure out the appropirate units. In this problem the units are suppressed for convenience. The suppressed units for x, v, a and t are m, m/s, m/s/s and sec, respectively.)


CHAPTER 2 MOTION ALONG A STRAIGHT LINE … · 12 CHAPTER 2 MOTION ALONG A STRAIGHT LINE Solutions to Exercises & Problems 1E ... 6E Calculate the time ... CHAPTER 2 MOTION ALONG A

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