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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% .