Download Problem: A person jogs eight complete laps around a quarter

PHYS 1301 - GENERAL PHYSICS I
Tuesday/Thursday 10:50AM-12:05PM - Dr. W. Anderson
Solutions to Homework from Chapter 2
Problem 7
Problem: A person jogs eight complete laps around a quarter-mile track in a total time of
12.5 min. Calculate (a) the average speed and (b) the average velocity, in m/s.
Solution: The total distance that the jogger covers is 8 lap · 0.25 mile/lap = 2 miles. However, the joggers total displacement is 0 miles, since she finishes at the same place she
starts. (a) Since speed is distance/time, the jogger’s speed is
1610 m 1 min
2.0 mi
·
·
= 4.3 m/s.
12.5 min
mi
60 s
(b) Since the jogger’s total displacement is 0.0 m, her average velocity is also 0.0 m.
Problem 16
Problem: A sports car is advertised to be able to stop in a distance of 50 m from a speed of
90 km/h. What is its acceleration in m/s2 ? How many g’s is this (1 g = 9.80 m/s2 )?
Solution: (a) We need to relate a displacement (50 m), a final velocity (0 m/s, since the car is
stopping) and an initial velocity (90 km/h) to an acceleration. The formula which does
that is
v 2 = v02 + 2a(x − x0 ).
First, let us rearrange this equation to isolate a. First, we subtract v02 from both sides,
so that we have
v 2 − v02 = v02 + 2a(x − x0 ) − v02 = 2a(x − x0 ).
Next, we want to divide both side by 2(x − x0 ),
2a(x − x0 )
v 2 − v02
=
= a.
2(x − x0 )
2(x − x0 )
Next, we need to express the initial velocity in m/s,
v0 = 90 km/h ·
1h
· 1000 m/km = 25 m/s.
3600 s
Finally, we can put v, v0 and x − x0 into our equation,
a=
(0 m/s)2 − (25 m/s)2
= 6.25 m/s2 .
2 · 50m
(b) To convert this to g’s, we simply multiply
a = 6.25 m/s2 ·
1
1g
0.64g.
9.80 m/s