fisica1-youn-e-freedman-exercicios-resolvidos

2.71: a) Travelling at 20 m s , Juan is x
1
= 37 m − (20 m s)(0.80 s) = 21m
from the
spreader when the brakes are applied, and the magnitude of the acceleration will be
2
v1
2x1
a = . Travelling at 25 m s , Juan is x = 37m − (25m s)(0.80s) 17m
from the
2
=
spreader, and the speed of the car (and Juan) at the collision is obtained from
v
2
x
= v
and so
2
0x
− 2a
x
x
2
= v
2
0x
v = 17.4 m s .
x
2
⎛ v ⎞
1
− 2
⎜
2
⎟ x
⎝ x1
⎠
2
= v
2
0x
− v
2
1
⎛ x
⎜
⎝ x
2
1
⎞
⎟ = (25 m s)
⎠
= 301 m
2
2
s
2
− (20 m s)
2
⎛17 m ⎞
⎜
21m
⎟
⎝ ⎠
b) The time is the reaction time plus the magnitude of the change in speed ( v 0
− v)
divided by the magnitude of the acceleration, or
2 v0
− v
25 m s −17.4 m s
t
flash
= treaction
+ x1
= (0.80 s) + 2
(21m) = 1.60 s.
2
v
(20 m s)
2
0
2.72: a) There are many ways to find the result using extensive algebra, but the most
straightforward way is to note that between the time the truck first passes the police car
and the time the police car catches up to the truck, both the truck and the car have
travelled the same distance in the same time, and hence have the same average velocity
over that time. Since the truck had initial speed 3
v and the average speed is v p , the
truck’s final speed must be 1
v .
2
p
2.73: a) The most direct way to find the time is to consider that the truck and the car are
initially moving at the same speed, and the time of the acceleration must be that which
gives a difference between the truck's position and the car's position as
24m + 21m + 26m + 4.5m = 75.5m,or t =
2
p
2
2(75.5m) (0.600 m s ) = 15.9s.
2
2
2
b) v t + (1 2) a t = (20.0 m s)(15.
9 s) + (1 2)(0.600 m s )(15.9 s) 394 m.
0 x
x
=
2
0 x
+ axt
= (20.0 m s) + (0.600 m s )(15.9 s) = 29.5 m
c) v
s.