1

Motion in One Dimension

SOLUTIONS TO PROBLEMS Section 2.1 Average Velocity

P2.3 (a) v =

!x!t

=10 m

2 s= 5 m s

(b) v = 5 m

4 s= 1.2 m s

(c) v =

x2 ! x1t2 ! t1

=5 m ! 10 m

4 s ! 2 s= !2.5 m s

(d) v =

x2 ! x1t2 ! t1

=!5 m ! 5 m

7 s ! 4 s= !3.3 m s

(e) v =

x2 ! x1t2 ! t1

=0 ! 08 ! 0

= 0 m s

*P2.4 (a) Let d represent the distance between A and B. Let t1 be the time for which the walker has

the higher speed in 5.00 m s = d

t1. Let t2 represent the longer time for the return trip in

!3.00 m s = ! d

t2. Then the times are

t1 =

d5.00 m s( )

and t2 =

d3.00 m s( )

. The average

speed is:

v = Total distanceTotal time

=d + d

d5.00 m s( ) +

d3.00 m s( )

=2d

8.00 m s( )d15.0 m2 s2( )

v =2 15.0 m2 s2( )

8.00 m s= 3.75 m s

(b) She starts and finishes at the same point A. With total displacement = 0, average velocity

= 0 .

2 Motion in One Dimension

Section 2.2 Instantaneous Velocity P2.5 (a) at ti = 1.5 s , xi = 8.0 m (Point A)

at t f = 4.0 s , x f = 2.0 m (Point B)

v =

x f ! xi

t f ! ti=

2.0 ! 8.0( ) m4 ! 1.5( ) s

= !6.0 m2.5 s

= !2.4 m s

(b) The slope of the tangent line can be found from Points C

and D. tC = 1.0 s, xC = 9.5 m( ) and tD = 3.5 s, xD = 0( ) ,

v ! "3.8 m s . (c) The velocity is zero when x is a minimum. This is at

t ! 4 s .

FIG. P2.5

P2.8 (a) v = 5 ! 0( ) m

1 ! 0( ) s= 5 m s

(b) v = 5 ! 10( ) m

4 ! 2( ) s= !2.5 m s

(c) v = 5 m ! 5 m( )

5 s ! 4 s( )= 0

(d) v = 0 ! !5 m( )

8 s ! 7 s( )= +5 m s

FIG. P2.8

Section 2.3 Analysis Models⎯The Particle Under Constant Velocity P2.9 Once it resumes the race, the hare will run for a time of

t =

x f ! xi

vx=

1 000 m ! 800 m8 m s

= 25 s .

In this time, the tortoise can crawl a distance

x f ! xi = 0.2 m s( ) 25 s( ) = 5.00 m .

Chapter 2 3

Section 2.4 Acceleration P2.12 (a) At t = 2.00 s ,

x = 3.00 2.00( )2 ! 2.00 2.00( ) + 3.00"

#$% m = 11.0 m .

At t = 3.00 s , x = 3.00 9.00( )2 ! 2.00 3.00( ) + 3.00"

#$% m = 24.0 m

so

v =

!x!t

=24.0 m " 11.0 m3.00 s " 2.00 s

= 13.0 m s .

(b) At all times the instantaneous velocity is

v = d

dt3.00t2 ! 2.00t + 3.00( ) = 6.00t ! 2.00( ) m s

At t = 2.00 s , v = 6.00 2.00( ) ! 2.00[ ] m s = 10.0 m s .

At t = 3.00 s , v = 6.00 3.00( ) ! 2.00[ ] m s = 16.0 m s .

(c) a = !v

!t=

16.0 m s " 10.0 m s3.00 s " 2.00 s

= 6.00 m s2

(d) At all times a = d

dt6.00t ! 2.00( ) = 6.00 m s2 . (This includes both t = 2.00 s and

t = 3.00 s ).

P2.13 x = 2.00 + 3.00t ! t2 , v =

dxdt

= 3.00 ! 2.00t , a = dv

dt= !2.00

At t = 3.00 s :

(a) x = 2.00 + 9.00 ! 9.00( ) m = 2.00 m (b) v = 3.00 ! 6.00( ) m s = !3.00 m s

(c) a = !2.00 m s2

Section 2.6 The Particle Under Constant Acceleration P2.21 (a) vi = 100 m s , a = !5.00 m s2 , v f = vi + at so 0 = 100 ! 5t , v f

2 = vi2 + 2a x f ! xi( ) so

0 = 100( )2 ! 2 5.00( ) x f ! 0( ) . Thus x f = 1 000 m and t = 20.0 s .

(b) At this acceleration the plane would overshoot the runway:

No .

4 Motion in One Dimension

P2.22 (a) Compare the position equation x = 2.00 + 3.00t ! 4.00t2 to the general form

x f = xi + vit +

12

at2

to recognize that xi = 2.00 m , vi = 3.00 m s , and a = !8.00 m s2 . The velocity equation,

v f = vi + at , is then

v f = 3.00 m s ! 8.00 m s2( )t .

The particle changes direction when v f = 0 , which occurs at t = 3

8 s . The position at this

time is:

x = 2.00 m + 3.00 m s( ) 3

8 s!

"#$%&' 4.00 m s2( ) 3

8 s!

"#$%&

2= 2.56 m .

(b) From x f = xi + vit +

12

at2 , observe that when x f = xi , the time is given by

t = ! 2vi

a. Thus,

when the particle returns to its initial position, the time is

t = !2 3.00 m s( )

!8.00 m s2 =34

s

and the velocity is v f = 3.00 m s ! 8.00 m s2( ) 3

4 s"

#$%&'= !3.00 m s .

Chapter 2 5

P2.25 (a) The time it takes the truck to reach 20.0 m s is found from v f = vi + at . Solving for t yields

t =

v f ! vi

a=

20.0 m s ! 0 m s2.00 m s2 = 10.0 s .

The total time is thus

10.0 s + 20.0 s + 5.00 s = 35.0 s . (b) The average velocity is the total distance traveled divided by the total time taken. The

distance traveled during the first 10.0 s is

x1 = vt =

0 + 20.02

!"#

$%&

10.0( ) = 100 m .

With a being 0 for this interval, the distance traveled during the next 20.0 s is

x2 = vit +

12

at2 = 20.0( ) 20.0( ) + 0 = 400 m .

The distance traveled in the last 5.00 s is

x3 = vt =

20.0 + 02

!"#

$%&

5.00( ) = 50.0 m .

The total distance x = x1 + x2 + x3 = 100 + 400 + 50 = 550 m , and the average velocity is

given by v =

xt=

55035.0

= 15.7 m s .

Section 2.7 Freely Falling Objects P2.29 (a) v f = vi ! gt : v f = 0 when t = 3.00 s , g = 9.80 m s2 . Therefore,

vi = gt = 9.80 m s2( ) 3.00 s( ) = 29.4 m s .

(b) y f ! yi =

12

v f + vi( )t

y f ! yi =

12

29.4 m s( ) 3.00 s( ) = 44.1 m

P2.31 (a) y f ! yi = vit +

12

at2 : 4.00 = 1.50( )vi ! 4.90( ) 1.50( )2 and vi = 10.0 m s upward .

(b) v f = vi + at = 10.0 ! 9.80( ) 1.50( ) = !4.68 m s

v f = 4.68 m s downward

6 Motion in One Dimension

P2.32 We have y f = !

12

gt2 + vit + yi

0 = ! 4.90 m s2( )t2 ! 8.00 m s( )t + 30.0 m .

Solving for t,

t = 8.00 ± 64.0 + 588

!9.80.

Using only the positive value for t, we find that t = 1.79 s .

*P2.34 (a) Consider the upward flight of the arrow.

vyf2 = vyi

2 + 2ay y f ! yi( )0 = 100 m s( )2 + 2 !9.8 m s2( )"y

"y = 10 000 m2 s2

19.6 m s2 = 510 m

(b) Consider the whole flight of the arrow.

y f = yi + vyit +12

ayt2

0 = 0 + 100 m s( )t + 12!9.8 m s2( )t2

The root t = 0 refers to the starting point. The time of flight is given by

t = 100 m s

4.9 m s2 = 20.4 s .

Additional Problems

P2.39 (a) a =

v f ! vi

t=

632 5 2803 600( )

1.40= !662 ft s2 = !202 m s2 = !20.6g

(b) x f = vit +

12

at2 = 632( ) 5 2803 600

!"#

$%&

1.40( ) ' 12

662( ) 1.40( )2 = 649 ft = 198 m

Chapter 2 7 P2.49 (a) Let x be the distance traveled at acceleration a until maximum speed v is reached. If this is

achieved in time t1 we can use the following three equations:

x =

12

v + vi( )t1 , 100 ! x = v 10.2 ! t1( ) and v = vi + at1 .

The first two give

100 = 10.2 ! 12

t1"#$

%&'

v = 10.2 ! 12

t1"#$

%&'

at1

a = 20020.4 ! t1( )t1

.

For Maggie: a = 20018.4( ) 2.00( )

= 5.43 m s2

For Judy: a = 20017.4( ) 3.00( )

= 3.83 m s2

(b) v = at1

Maggie: v = 5.43( ) 2.00( ) = 10.9 m s

Judy: v = 3.83( ) 3.00( ) = 11.5 m s

(c) At the six-second mark

x =

12

at12 + v 6.00 ! t1( )

Maggie: x = 12

5.43( ) 2.00( )2 + 10.9( ) 4.00( ) = 54.3 m

Judy: x = 12

3.83( ) 3.00( )2 + 11.5( ) 3.00( ) = 51.7 m

Maggie is ahead by 2.62 m .

P2.55 (a) d = 1

29.80( )t1

2 d = 336t2

t1 + t2 = 2.40 336t2 = 4.90 2.40 ! t2( )2

4.90t22 ! 359.5t2 + 28.22 = 0

t2 =

359.5 ± 359.52 ! 4 4.90( ) 28.22( )9.80

t2 =

359.5 ± 358.759.80

= 0.076 5 s so d = 336t2 = 26.4 m

(b) Ignoring the sound travel time, d = 1

29.80( ) 2.40( )2 = 28.2 m , an error of 6.82% .