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m _
2.9. Note that the two speeds given (100.0 km/h and 50.0 km/h) are _
5.00

s

average speeds for two different legs of a trip. They are not initial = m _

and final speeds of an accelerating object, so you cannot add 3.0
2
s

them and divide by 2. The average speed for the total (entire)
5.00 m _ _
s·s
trip can be found from the definition of average speed; that is, = _ ×

3.0 s m
average speed is the total distance covered divided by the total

time elapsed. So, we start by finding the distance covered for = 1.7 s
−
each of the two legs of the trip: 2.12. The relationship between average velocity ( v ), distance (d), and

− _ v − t time (t) can be solved for time:
d
v = ∴ d =
t
− _
d
v =

km
( _ ) t

Leg 1 distance = 100.0 (2.00 h)
h − v t = d
= 200.0 km d _
t =

−
v
( _ )
km

Leg 2 distance = 50.0 (1.00 h)
h _
1,609 m
t =

= 50.0 km 720 m/s
1,609 m _ _
s
Total distance (leg 1 plus leg 2) = 250.0 km = _ ×

Total time = 3.00 h 720 1 m
_
m·s
− _ _ = 2.23
250.0 km
d
m

v = = = 83.3 km/h

t 3.00 h
= 2.2 s
2.10. The initial velocity, final velocity, and time are known, and the −

2.13. The relationship between average velocity ( v ), distance (d), and

problem asked for the acceleration. Listing these quantities with
time (t) can be solved for distance:
their symbols, we have
− _ −
d

v
v i = 0 v = ∴ d = t
t
v f = 15.0 m/s m _
d = ( 40.0 ) (0.4625 s)
s
t = 10.0 s
m·s
_
a = ? = 40.0 × 0.4625

s

These are the quantities involved in the acceleration equation, = 18.5 m
which is already solved for the unknown:
2.14. “How many minutes . . .” is a question about time, and the
_ distance is given. Since the distance is given in km and the speed
v f – v i
a =

t in m/s, a unit conversion is needed. The easiest thing to do is to

__ convert km to m. There is 1,000 m in 1 km, and
15.0 m/s – 0 m/s
a =
10.0 s 8 3 11
(1.50 × 10 km) × (1 × 10 m/km) = 1.50 × 10 m
_ 1
15.0 m _ _

= × −

v

10.0 s s The relationship between average velocity ( ), distance (d), and
time (t) can be solved for time:
m _
= 1.50
d
s 2 − _ d _
v = ∴ t =

t − v

2.11. The initial velocity, final velocity, and acceleration are known,

11
1.50 × 10 m
__
and the problem asked for the time. Listing these quantities t = 8 m _

with their symbols, we have 3.00 × 10
s
s
1.50
v i = 20.0 m/s = _ × 10 11–8 m _ _
×

3.00 1 m
v f = 25.0 m/s
_
3 m·s

a = 3.0 m/s 2 = 0.500 × 10
m
t = ? = 5.00 × 10 s
2
500
These are the quantities involved in the acceleration equation, _ _ s _ _
min
500 s

=

×

s _
which first must be solved for the unknown time: 60 60 1 s

min
_ _ _
v f – v i
v f – v i
s·min

a = ∴ t = = 8.33
t a s
m _ m _ = 8.33 min

25.0 – 20.0

__
s
s
t = (Information on how to use scientific notation [also called
m _

3.0

s 2 powers of ten or exponential notation] is located in appendix A,
Mathematical Review.)
646 APPENDIX E Solutions for Group A Parallel Exercises E-4