fisica1-youn-e-freedman-exercicios-resolvidos

3.79: Finding the infinite series consisting of the times between meeting with the brothers
is possible, and even entertaining, but hardly necessary. The relative speed of the brothers
is 70 km/h, and as they are initially 42 km apart, they will reach each other in six-tenths
of an hour, during which time the pigeon flies 30 km.
3.80: a) The drops are given as falling vertically, so their horizontal component of
velocity with respect to the earth is zero. With respect to the train, their horizontal
component of velocity is 12.0 m/s, west (as the train is moving eastward). b) The
vertical component, in either frame, is ( 12 .0 m/s)
/(
tan30° ) = 20.
8 m/s,
and this is the
magnitude of the velocity in the frame of the earth. The magnitude of the velocity in the
frame of the train is
( 12.0 m/s) /sin30°.
(12.0 m/s)
2
2
+ ( 20.
8 m/s) = 24 m/s. This is, of course, the same as
3.81: a) With no wind, the plane would be 110 km west of the starting point; the wind has
blown the plane 10 km west and 20 km south in half an hour, so the wind velocity is
2
2
(20 km/h) + (40 km/h) = 44.7 km/h at a direction of arctan( 40/ 20) = 63°
south of
west. b) arcsin( 40/ 220) = 10. 5°
north of west.
2 2
3.82: a) 2 D / v b) 2Dv /( v − w ) c)
2 2
D / v − d) 1.50 h, 1.60 h, 1.55 h.
2 w
2
3.83: a) The position of the bolt is 3.00 m + (2.50 m/s) t −1/
2(9.80 m/s 2 ) t , and the
position of the floor is (2.50 m/s)t. Equating the two, 3.00 m = (4.90 m/s
2 ) t 2 . Therefore
2
t = 0.782 s. b) The velocity of the bolt is 2.50 m/s − (9.80 m/s )(0.782 s) = −5.17
m/s
relative to Earth, therefore, relative to an observer in the elevator
v = −5.17 m/s − 2.50 m/s = −7.67 m/s. c) As calculated in part (b), the speed relative to
Earth is 5.17 m/s. d) Relative to Earth, the distance the bolt travelled is
2 2
2
2
(2.50 m/s) t − 1/ 2(9.80 m/s ) t = (2.50 m/s)(0.782 s)
− ( 4.90 m/s )(0.782 s) = −1.04 m
5310 km
3.84: Air speed of plane = = 804.5 km/h
With wind from A to B:
Same distance both ways:
6.60 h
AB
+ tBA
=
t
6.70 h
5310 km
804.5 km/h + v
w
) t = = 2655 km
2
804.5 km/h + v )t 2655 km
(
AB
(
w BA
=
Solve (1), (2), and (3) to obtain wind speed v
w
:
v = 98.1km/h
w