Question

A train starts from rest and moves with a constant acceleration of 2.0 $\frac{m}{s^2}$ fbr half a minute. The brakes are then applied and the train comes to rest in one minute. Find.

(a) the total distance moved. by the train,

(b) the maximum speed attained by the train and

(c) the position(s) of the train at half the maxim um speed

Answer 1

Intial velocity, u = 0

Acceleration, a = 2$\frac{m}{s^2}$

Let final velocity be v

(before appling breaks)

t = 30 sec

v = u + at

=0 + 2$\times 30$

=60 $\frac{m}{s}$

(a) $S_1$ = ut + $\frac{1}{2}a{t}^2$
=$\frac{1}{2}\times 2\times (30{)}^2$
= 900,

When breaks are applied,

u' = 60 $\frac{m}{s}$

v' = 0

t = 60 sec (1 min)

Deceleration,

$a^1=\frac{v-u}{t}$
= $\frac{0-60}{60}=-1\frac{m}{s^2}$
$S_2$ = $\frac{v^2-u^2}{2a^{\prime }}$
=$\frac{0^2-{60}^2}{2(-1)}$
= 1800 m

Total S = $S_1+S_2$
=1800+900
=2700 m
=2.7 km

Tha maximum speed attained by train, v= 60$\frac{m}{s}$

Half the maximum speed

= $\frac{60}{2}=30\frac{m}{s}$
Distance, S =$\frac{v^2-u^2}{2a^{\prime }}$
= 1350 m
Position = 900-1350
=2250
=2.25 km
from stating point