fisica1-youn-e-freedman-exercicios-resolvidos

2.68: One convenient way to do the problem is to do part (b) first; the time spent
20 m s
accelerating from rest to the maximum speed is = 80 s.
At this time, the officer is
2
v1
x =
2 a
(20 m s)
2
1
=
=
2
2(2.5 m s
)
2
2.5 m s
80.0 m.
This could also be found from (1/ 2) a t 2
1 1
, where t 1 is the time found for the acceleration.
At this time the car has moved (15 m s )(8.0 s) = 120 m, so the officer is 40 m behind the
car.
a) The remaining distance to be covered is 300 m – x 1 and the average speed is
(1/2)(v 1 + v 2 ) = 17.5 m s , so the time needed to slow down is
360 m − 80 m
17.5 m s
and the total time is 24.0 s.
= 16.0 s,
c) The officer slows from 20 m s to 15 m s in 16.0 s (the time found in part (a)), so
the acceleration is –0.31
d), e)
2
m s .