108 Aptitude Test Problems in Physics
time of ascent irrespective of the type of this
force. Indeed, if in the process of ascent the body
attains an intermediate height h', its velocity v'
at this point, required to reach the height h in
the presence of drag, must be higher than the ve-
locity in the absence of drag since a fraction of the
kinetic energy will be transformed into heat during
the subsequent ascent. The body sliding down
from the height h and reaching the height h' will
have (due to the work done by the drag force) a
velocity v" which is lower than the velocity of the
body moving down without a dray. Thus, while
passing by the same point on the inclined plane,
the ascending body has a higher veloeity than the
descending body. For this reason, the ascending
body will cover a small distance in the vicinity of
point h' in a shorter time than the descending body.
Dividing the entire path into small regions, we
see that each region will be traversed by the ascend-
ing body in a shorter time than by the descend-
ing body. Consequently, the total time of ascent
will be shorter than the time of descent.
1.2. Since the locomotive moves with a constant
deceleration after the application of brakes, it will
come to rest in t = vl a = 50 s, during which it will
cover a distance s = v 2 /(2a) = 375 m. Thus, in
1 min after the application of brakes, the loco-
motive will be at a distance 1 = L — s = 25 m
from the traffic light.
1.3. At the moment the pilot switches off the en-
gine, the helicopter is at an altitude h
Since the sound can no longer be heard on the
ground after a time t 2 , we obtain the equation
at;
t 2 = ti -F ,where on the right-hand side we have the time of
ascent of the helicopter to the altitude h and the
time taken for the sound to reach the ground from
the altitude h. Solving the obtained quadratic
equation, we find that3 ( ) 2 +
C
12 -
a a