fisica1-youn-e-freedman-exercicios-resolvidos

2.66: a) The simplest way to do this is to go to a frame in which the freight train (which
moves with constant velocity) is stationary. Then, the passenger train has
an initial relative velocity of v rel,0 = 10 m s . This relative speed would be decreased to
vrel
,0
zero after the relative separation had decreased to = + 500 m.
Since this is larger in
magnitude than the original relative separation of 200 m, there will be a collision. b) The
time at which the relative separation goes to zero (i.e., the collision time) is found by
solving a quadratic (see Problems 2.35 & 2.36 or Example 2.8). The time is given by
2
2arel
2
( vrel,0
− vrel,0
2axrel,0
)
1
t =
+
a
= (10 s
2
m)(10 m s −
((1
0.6)
= ( 100 s) − .
100 m
2
s
2
− 40 m
Substitution of this time into Eq. (2.12), with x 0 = 0, yields 538 m as the distance the
passenger train moves before the collision.
2
s
2
)
2.67: The total distance you cover is 1 .20 m + 0.90 m = 2.10 m and the time available is
1
= 0.80 s . Solving Eq. (2.12) for a x ,
.20 m
1.50 m s
( x − x ) − v
t
t
(2.10 m) − (0.80m s)(0.80 s)
(0.
80 s)
0 0x
s
2
x
= 2 = 2
= 4.56 m .
2
2
a