Question

The driver of a train travelling at 40m/s applies the breaks as the train enters a station. The train slows down at a rate 2$m{{s}^{-2}}$. The platform is 400m long. Will the train stop in time?

Answer 1

Hint: After applying the breaks, the train will undergo retardation. Hence, it will come to rest after some time. For the train to reach in time, it covers a distance of 400m after the brakes are applied. Find its displacement from one of the kinematic equations.
Formula used:
$2as={{v}^{2}}-{{u}^{2}}$

Complete answer:
It is given that the train slows down as it approaches a station. It is given that the train slows down at a rate of 2$m{{s}^{-2}}$. This means that when the driver applies the breaks, the train undergoes retardation.
It is given that the initial velocity of the train is 40m/s. And the length of the platform is given to be 400m. Therefore, for the train to stop in time, it must come to rest with a displacement of 400m from the time when the brakes are applied.
Let us find the displacement of the train. For thus we will use the kinematic equation $2as={{v}^{2}}-{{u}^{2}}$ …… (i),
a is the acceleration of the train, s is the displacement, v id the final velocity and u is the initial velocity.
Here, a = -2$m{{s}^{-2}}$, v = 0, u = 40$m{{s}^{-1}}$.
Substitute the values in equation (i)
$\Rightarrow 2(-2)s={{0}^{2}}-{{(40)}^{2}}$.
$\Rightarrow 4s=1600$
$\Rightarrow s=400m$.
This means that it moves further by 400m after the brakes are applied and the length of the platform is also 400m. Therefore, the train will reach in time.

Note:
Do not confuse between acceleration and retardation.
Actually, retardation is an acceleration. When we say that a body is retarding, it means that its velocity is decreasing with time and therefore, it is accelerated in the opposite direction of its velocity.