136 Aptitude Test Problems in Physics
load (and the carriage) is determined by this
horizontal force: ma=mglia 2 — 1. Consequently,
a — 1 = (3/4)g = 7.5 m/s 2.
Fig. 146
On the first segment of the path, the carriage is
accelerated to the velocity v = ati = (3/4) gt i =
30 m/s and covers the distance si in the forward
direction in the straight line:
ail 3
= gt-
a
= 60 m.
2 8 I
Further, it moves at a constant velocity v during
the time interval t 2 = 3 s and traverses the path
of length
ss = vts = 90 m.
Thus, seven seconds after the beginning of motion,
the carriage is at a distance si ss = 150 m in
front of the initial position.
On the third segment, the carriage moves round
a bend to the right. Since the velocity of the car-
riage moving on the rails is always directed along
the carriage, the constant (during the time interval
is = 25.12 s) transverse acceleration a = (3/4) g
is a centripetal acceleration, i.e. the carriage moves
in a circle at a constant velocity v: a = v 2 /R, the
radius of the circle being R = v 2 /a = 120 m. The
path traversed by the carriage in the circle is
as = Rp = vts ,
Si =