Solutions
d x
3-1. (a) From v = ⇒ v= .
t t
x
(b) v = . We want the answer in m/s so we’ll need to convert 30.0 km to meters and 8.0
t
min to seconds: 30.0 km ! 1000 1 km = 30, 000 m; 8.0 min! 1 min = 480 s.
m 60 s
x 30,000 m
Then v = = = 63 ms .
t 480 s
Alternatively, we can do the conversions within the equation:
x 30.0 km ! 1000 m
v= = 1 km
= 63 ms .
t 8.0 min! 160mins
In mi/h:
x 18.6 mi
30.0 km ! 1.61 km = 18.6 mi; 8.0 min! 60 min = 0.133 h. Then v =
1 mi 1h
= = 140 mi
h . Or,
t 0.133 h
x 30.0 km 60 min 1 mi x 30.0 km ! 1.611 mi
v= = ! ! = 140 mih . Or, v = = km
= 140 mi
h .
t 8.0 min 1h 1.61 km t 8.0 min! 601min
h
There is usually more than one way to approach a problem and arrive at the correct answer!
d L
3-2. (a) From v = ⇒ v= .
t t
L 24.0m
(b) v = = = 40 s .
m
t 0.60s
d x
3-3. (a) From v = ⇒ v= .
t t
x 0.30m
(b) v = = = 30 ms .
t 0.010s
d 2 !r
3-4. (a) v = = .
t t
2! r 2! (400m)
(b) v = = = 63 ms .
t 40s
d h
3-5. (a) t = ? From v = !t = .
t v
h 508 m
(b) t = = = 34 s.
v 15 ms
(c) Yes. At the beginning of the ride the elevator has to speed up from rest, and at the end of
the ride the elevator has to slow down. These slower portions of the ride produce an
average speed lower than the peak speed.
© Paul G. Hewitt and Phillip R. Wolf 3-1
,3-6. (a) t = ? Begin by getting consistent units. Convert 100.0 yards to meters using the
conversion factor on the inside cover of your textbook: 0.3048 m = 1.00 ft.
3ft
( ) (
0.3048m
)
Then 100.0 yards ! 1 yard ! 1 ft = 91.4 m. From v =
d
t
!t = =
d 91.4 m
v v
.
d 91.4 m
(b) t = = = 15 s.
v ( 6.0 ms )
d d L
3-7. (a) t = ? From v = !t= = .
t v c
L 1.00 m
(b) t = = = 3.33 ! 10 -9 s = 3.33 ns. (This is 3 13 billionths of a second!)
c 3.00 ! 10 s
8 m
d
3-8. (a) d = ? From v = ! d = vt.
t
(b) First, we need a consistent set of units. Since speed is in m/s let’s convert minutes to seconds:
5.0 min ! 1min = 300 s. Then d = v t = 7.5 ms ! 300s = 2300 m.
60s
v0 + vf 0 + v v
3-9. (a) v = = = .
2 2 2
d vt
(b) d = ? From v = ! d = vt = .
t 2
vt ( 2.0 s )
m
(1.5s)
(c) d = = = 1.5 m.
2 2
d "v +v % " 0 + v% vt
3-10. (a) d = ? From v = ! d = vt = $ 0 f ' t = $ ' t= .
t # 2 & # 2 & 2
vt (12 s )(8.0s)
m
(b) d = = = 48 m.
2 2
d "v +v % " 0 + v% vt
3-11. (a) d = ? From v = ! d = vt = $ 0 f ' t = $ ' t= .
t # 2 & # 2 & 2
(b) First get consistent units: 100.0 km/h should be expressed in m/s (since the time is in
vt ( 27.8 ms ) (8.0 s)
seconds). 100.0 km h ! ( 3600 s ) ! ( 1 km ) = 27.8 s . Then, d = = = 110 m.
1h 1000 m m
2 2
!v v 2 " v1
3-12. (a) a = = .
!t t
(b) !v = 40 km h " 15 h = 25 h . Since our time is in seconds we need to convert h to s :
km km km m
"v 6.94 ms
25 kmh ! 3600 s ! 1 km = 6.94 s . Then a = = = 0.35 sm2 .
1hr 1000 m m
t 20 s
Alternatively, we can express the speeds in m/s first and then do the calculation:
11.1 ms " 4.17 ms
15 km
hr ! 1hr
3600 s ! 1000 m
1 km = 4.17 m
s and 40 km
hr ! 1hr
3600 s ! 1000 m
1 km = 11.1 m
s . Then a = = 0.35 sm2 .
20s
© Paul G. Hewitt and Phillip R. Wolf 3-2
, !v v 2 " v1
3-13. (a) a = = .
!t t
(b) To make the speed units consistent with the time unit we’ll need Δv in m/s:
v2 " v1 4.17 ms
!v = v2 " v1 = 20.0 km h " 5.0 h = 15.0 h # 3600 s # 1 km = 4.17 s . Then a =
km km 1hr 1000 m m
= = 0.417 sm2 .
t 10.0 s
An alternative is to convert the speeds to m/s first:
v1 = 5.0 km
h ! 3600 s ! 1 km = 1.39 s ;
1hr 1000 m m
v2 = 20.0 kmh ! 3600 s ! 1 km = 5.56 s .
1hr 1000 m m
v2 ! v1 ( 5.56 ms ! 1.39 ms )
Then a = = = 0.417 sm2 .
t 10.0 s
v1 + v2 ! 1.39 ms + 5.56 ms $
(c) d = vt = t=# &% 10.0 s = 35 m. Or,
2 " 2
( )
d = v1t + 12 at 2 = (1.39 ms ) (10.0 s) + 12 0.417 sm2 (10.0 s)2 = 35 m.
There is almost always more than one way to solve a physics problem correctly!
!v v f " v 0 0 " v "v
3-14. (a) a = = = = .
!t !t t t
!v !25 ms
(b) a = = = 1.25 sm2 = !1.3 sm2 (to 2 significant figures). .
t 20s
d "v +v % " 25 ms + 0 ms %
(c) d = ? From v = ! d = vt = $ 0 f ' t = $ '& 20 s = 250 m.
t # 2 & # 2
(Or, d = v t + 0
1
2 ( ) )
at 2 = 25 ms (20 s) + 12 !1.25 sm2 (20 s)2 = 250 m.
(d) d = ? Lonnie travels at a constant speed of 25 m/s before applying the brakes, so
d = vt = ( 25 ms ) (1.5 s) = 38 m.
!v v f " v 0 0 " v "v
3-15. (a) a = = = = .
!t !t t t
!v !72 m
(b) a = = s
= !6.0 m2 .
t 12 s s
d "v +v % " 72 ms + 0 ms %
(c) d = ? From v = ! d = vt = $ 0 f ' t = $ '& (12 s) = 430 m.
t # 2 & # 2
( )
Or, d = v0 t + 12 at 2 = 72 ms (12 s) + 12 !6.0 sm2 (12 s)2 = 430 m.
d d L 2L
3-16. (a) t = ? From v = !t = = =
( )
vf +v0
.
t v 2
v
2L 2(1.4m)
(b) t = = = 0.19 s.
v 15.0 ms
© Paul G. Hewitt and Phillip R. Wolf 3-3
, !v +v $ v
3-17. (a) v = # 0 f & = .
" 2 % 2
350 ms
(b) v = = 175 ms . Note that the length of the barrel isn’t needed—yet!
2
d d L 0.40 m
(c) From v = ! t= = = = 0.0023 s.
t v v 175 ms
d "v +v % " v + v%
3-18. (a) From v = ! d = vt = $ 0 f ' t = $ 0
# 2 '&
t.
t # 2 &
! v + v$ ! 25 ms + 11 ms $
(b) d = # 0 & t =# &% (8.0s) = 140 m.
" 2 % " 2
= ( 24 1s ) x. (That’s 24x per
d x
3-19. (a) v = ? There’s a time t between frames of 1
s, so v= =
24
t ( 1
24 s)
second.)
(b) v = ( 24 1s ) x = ( 24 1s ) (0.40 m) = 9.6 ms .
3-20. (a) a = ? Since time is not a part of the problem we can use the formula v f ! v 0 = 2ad and
2 2
v2
solve for acceleration a. Then, with v0 = 0 and d = x, a = .
2x
( )
2
v2 1.8 ! 10 7 ms
(b) a = = = 1.6 ! 1015 sm2 .
2x 2(0.10 m)
(c) t = ? From vf = v0 + at !t =
vf " v0
=
(
1.8 # 10 7 ms " 0 ms
= 1.1 # 10 -8 s.
)
a 1.6 # 1015 sm2
# d d L 2L 2(0.10 m) &
% Or, from v = ! t= = = = = " -8
(
( v +v2 ) ( )
1.1 10 s.
%$ t v f 0 (v + 0) 1.8 " 10 7 ms ('
d "v +v % " v+V%
3-21. (a) d=? From v = ! d = vt = $ 0 f ' t = $
# 2 '&
t.
t # 2 &
! v +V $ ! 110 ms + 250 ms $
(b) d = # &% (3.5s) = 630 m.
" 2 &% #"
t =
2
3-22. (a) t = ? Let's choose upward to be the positive direction.
v !v 0!v v
From vf = v0 + at with vf = 0 and a = !g " t = f 0 = = .
a !g g
m
v 32 s
(b) t = = = 3.3 s.
g 9.8 sm2
d "v +v % " v% " v% v
(c) d = ? From v = ! d = vt = $ 0 f ' t = $ ' $ ' =
2
=
( 32 ms ) = 52 m. 2
t # 2 & # 2 & # g & 2g 2 9.8 m2
s ( )
We get the same result with d = v0 t + 12 at 2 = ( 32 ms ) (3.3 s) + 12 !9.8 sm2 (3.3 s)2 = 52 m.( )
© Paul G. Hewitt and Phillip R. Wolf 3-4
d x
3-1. (a) From v = ⇒ v= .
t t
x
(b) v = . We want the answer in m/s so we’ll need to convert 30.0 km to meters and 8.0
t
min to seconds: 30.0 km ! 1000 1 km = 30, 000 m; 8.0 min! 1 min = 480 s.
m 60 s
x 30,000 m
Then v = = = 63 ms .
t 480 s
Alternatively, we can do the conversions within the equation:
x 30.0 km ! 1000 m
v= = 1 km
= 63 ms .
t 8.0 min! 160mins
In mi/h:
x 18.6 mi
30.0 km ! 1.61 km = 18.6 mi; 8.0 min! 60 min = 0.133 h. Then v =
1 mi 1h
= = 140 mi
h . Or,
t 0.133 h
x 30.0 km 60 min 1 mi x 30.0 km ! 1.611 mi
v= = ! ! = 140 mih . Or, v = = km
= 140 mi
h .
t 8.0 min 1h 1.61 km t 8.0 min! 601min
h
There is usually more than one way to approach a problem and arrive at the correct answer!
d L
3-2. (a) From v = ⇒ v= .
t t
L 24.0m
(b) v = = = 40 s .
m
t 0.60s
d x
3-3. (a) From v = ⇒ v= .
t t
x 0.30m
(b) v = = = 30 ms .
t 0.010s
d 2 !r
3-4. (a) v = = .
t t
2! r 2! (400m)
(b) v = = = 63 ms .
t 40s
d h
3-5. (a) t = ? From v = !t = .
t v
h 508 m
(b) t = = = 34 s.
v 15 ms
(c) Yes. At the beginning of the ride the elevator has to speed up from rest, and at the end of
the ride the elevator has to slow down. These slower portions of the ride produce an
average speed lower than the peak speed.
© Paul G. Hewitt and Phillip R. Wolf 3-1
,3-6. (a) t = ? Begin by getting consistent units. Convert 100.0 yards to meters using the
conversion factor on the inside cover of your textbook: 0.3048 m = 1.00 ft.
3ft
( ) (
0.3048m
)
Then 100.0 yards ! 1 yard ! 1 ft = 91.4 m. From v =
d
t
!t = =
d 91.4 m
v v
.
d 91.4 m
(b) t = = = 15 s.
v ( 6.0 ms )
d d L
3-7. (a) t = ? From v = !t= = .
t v c
L 1.00 m
(b) t = = = 3.33 ! 10 -9 s = 3.33 ns. (This is 3 13 billionths of a second!)
c 3.00 ! 10 s
8 m
d
3-8. (a) d = ? From v = ! d = vt.
t
(b) First, we need a consistent set of units. Since speed is in m/s let’s convert minutes to seconds:
5.0 min ! 1min = 300 s. Then d = v t = 7.5 ms ! 300s = 2300 m.
60s
v0 + vf 0 + v v
3-9. (a) v = = = .
2 2 2
d vt
(b) d = ? From v = ! d = vt = .
t 2
vt ( 2.0 s )
m
(1.5s)
(c) d = = = 1.5 m.
2 2
d "v +v % " 0 + v% vt
3-10. (a) d = ? From v = ! d = vt = $ 0 f ' t = $ ' t= .
t # 2 & # 2 & 2
vt (12 s )(8.0s)
m
(b) d = = = 48 m.
2 2
d "v +v % " 0 + v% vt
3-11. (a) d = ? From v = ! d = vt = $ 0 f ' t = $ ' t= .
t # 2 & # 2 & 2
(b) First get consistent units: 100.0 km/h should be expressed in m/s (since the time is in
vt ( 27.8 ms ) (8.0 s)
seconds). 100.0 km h ! ( 3600 s ) ! ( 1 km ) = 27.8 s . Then, d = = = 110 m.
1h 1000 m m
2 2
!v v 2 " v1
3-12. (a) a = = .
!t t
(b) !v = 40 km h " 15 h = 25 h . Since our time is in seconds we need to convert h to s :
km km km m
"v 6.94 ms
25 kmh ! 3600 s ! 1 km = 6.94 s . Then a = = = 0.35 sm2 .
1hr 1000 m m
t 20 s
Alternatively, we can express the speeds in m/s first and then do the calculation:
11.1 ms " 4.17 ms
15 km
hr ! 1hr
3600 s ! 1000 m
1 km = 4.17 m
s and 40 km
hr ! 1hr
3600 s ! 1000 m
1 km = 11.1 m
s . Then a = = 0.35 sm2 .
20s
© Paul G. Hewitt and Phillip R. Wolf 3-2
, !v v 2 " v1
3-13. (a) a = = .
!t t
(b) To make the speed units consistent with the time unit we’ll need Δv in m/s:
v2 " v1 4.17 ms
!v = v2 " v1 = 20.0 km h " 5.0 h = 15.0 h # 3600 s # 1 km = 4.17 s . Then a =
km km 1hr 1000 m m
= = 0.417 sm2 .
t 10.0 s
An alternative is to convert the speeds to m/s first:
v1 = 5.0 km
h ! 3600 s ! 1 km = 1.39 s ;
1hr 1000 m m
v2 = 20.0 kmh ! 3600 s ! 1 km = 5.56 s .
1hr 1000 m m
v2 ! v1 ( 5.56 ms ! 1.39 ms )
Then a = = = 0.417 sm2 .
t 10.0 s
v1 + v2 ! 1.39 ms + 5.56 ms $
(c) d = vt = t=# &% 10.0 s = 35 m. Or,
2 " 2
( )
d = v1t + 12 at 2 = (1.39 ms ) (10.0 s) + 12 0.417 sm2 (10.0 s)2 = 35 m.
There is almost always more than one way to solve a physics problem correctly!
!v v f " v 0 0 " v "v
3-14. (a) a = = = = .
!t !t t t
!v !25 ms
(b) a = = = 1.25 sm2 = !1.3 sm2 (to 2 significant figures). .
t 20s
d "v +v % " 25 ms + 0 ms %
(c) d = ? From v = ! d = vt = $ 0 f ' t = $ '& 20 s = 250 m.
t # 2 & # 2
(Or, d = v t + 0
1
2 ( ) )
at 2 = 25 ms (20 s) + 12 !1.25 sm2 (20 s)2 = 250 m.
(d) d = ? Lonnie travels at a constant speed of 25 m/s before applying the brakes, so
d = vt = ( 25 ms ) (1.5 s) = 38 m.
!v v f " v 0 0 " v "v
3-15. (a) a = = = = .
!t !t t t
!v !72 m
(b) a = = s
= !6.0 m2 .
t 12 s s
d "v +v % " 72 ms + 0 ms %
(c) d = ? From v = ! d = vt = $ 0 f ' t = $ '& (12 s) = 430 m.
t # 2 & # 2
( )
Or, d = v0 t + 12 at 2 = 72 ms (12 s) + 12 !6.0 sm2 (12 s)2 = 430 m.
d d L 2L
3-16. (a) t = ? From v = !t = = =
( )
vf +v0
.
t v 2
v
2L 2(1.4m)
(b) t = = = 0.19 s.
v 15.0 ms
© Paul G. Hewitt and Phillip R. Wolf 3-3
, !v +v $ v
3-17. (a) v = # 0 f & = .
" 2 % 2
350 ms
(b) v = = 175 ms . Note that the length of the barrel isn’t needed—yet!
2
d d L 0.40 m
(c) From v = ! t= = = = 0.0023 s.
t v v 175 ms
d "v +v % " v + v%
3-18. (a) From v = ! d = vt = $ 0 f ' t = $ 0
# 2 '&
t.
t # 2 &
! v + v$ ! 25 ms + 11 ms $
(b) d = # 0 & t =# &% (8.0s) = 140 m.
" 2 % " 2
= ( 24 1s ) x. (That’s 24x per
d x
3-19. (a) v = ? There’s a time t between frames of 1
s, so v= =
24
t ( 1
24 s)
second.)
(b) v = ( 24 1s ) x = ( 24 1s ) (0.40 m) = 9.6 ms .
3-20. (a) a = ? Since time is not a part of the problem we can use the formula v f ! v 0 = 2ad and
2 2
v2
solve for acceleration a. Then, with v0 = 0 and d = x, a = .
2x
( )
2
v2 1.8 ! 10 7 ms
(b) a = = = 1.6 ! 1015 sm2 .
2x 2(0.10 m)
(c) t = ? From vf = v0 + at !t =
vf " v0
=
(
1.8 # 10 7 ms " 0 ms
= 1.1 # 10 -8 s.
)
a 1.6 # 1015 sm2
# d d L 2L 2(0.10 m) &
% Or, from v = ! t= = = = = " -8
(
( v +v2 ) ( )
1.1 10 s.
%$ t v f 0 (v + 0) 1.8 " 10 7 ms ('
d "v +v % " v+V%
3-21. (a) d=? From v = ! d = vt = $ 0 f ' t = $
# 2 '&
t.
t # 2 &
! v +V $ ! 110 ms + 250 ms $
(b) d = # &% (3.5s) = 630 m.
" 2 &% #"
t =
2
3-22. (a) t = ? Let's choose upward to be the positive direction.
v !v 0!v v
From vf = v0 + at with vf = 0 and a = !g " t = f 0 = = .
a !g g
m
v 32 s
(b) t = = = 3.3 s.
g 9.8 sm2
d "v +v % " v% " v% v
(c) d = ? From v = ! d = vt = $ 0 f ' t = $ ' $ ' =
2
=
( 32 ms ) = 52 m. 2
t # 2 & # 2 & # g & 2g 2 9.8 m2
s ( )
We get the same result with d = v0 t + 12 at 2 = ( 32 ms ) (3.3 s) + 12 !9.8 sm2 (3.3 s)2 = 52 m.( )
© Paul G. Hewitt and Phillip R. Wolf 3-4
