fisica1-youn-e-freedman-exercicios-resolvidos

2.31: a) At t = 3 s the graph is horizontal and the acceleration is 0. From t = 5 s to t =
45 m s−20m s
2
9 s, the acceleration is constant (from the graph) and equal to = 6.3 m s . From
t = 9 s to t = 13 s the acceleration is constant and equal to
0−45 m s
2
= −11.2
m s .
4 s
b) In the first five seconds, the area under the graph is the area of the rectangle, (20
m)(5 s) = 100 m. Between t = 5 s and t = 9 s, the area under the trapezoid is (1/2)(45 m/s
+ 20 m/s)(4 s) = 130 m (compare to Eq. (2.14)), and so the total distance in the first 9 s is
230 m. Between t = 9 s and t = 13 s, the area under the triangle is
( 1 2)(45 m s)(4 s) = 90 m , and so the total distance in the first 13 s is 320 m.
4 s
2.32:
2.33: a) The maximum speed will be that after the first 10 min (or 600 s), at which time
the speed will be
2
(20.0 m s )(900 s) = 1.8×
10
4
m s = 18 km s.
b) During the first 15 minutes (and also during the last 15 minutes), the ship will travel
( 1 2)(18 km s)(900 s) = 8100 km , so the distance traveled at non-constant speed is 16,200
km and the fraction of the distance traveled at constant speed is
16,200 km
1 −
= 0.958,
384,000 km
keeping an extra significant figure.
384,000 km−16,200 km
4
c) The time spent at constant speed is = 2.04×
10 s and the time spent
18 km s
during both the period of acceleration and deceleration is 900 s, so the total time required
4
for the trip is 2.22× 10 s , about 6.2 hr.