Question

A train is crossing an observer standing on a platform. The first compartment of the train takes 2 second to cross the observer while the second compartment takes 2.5 seconds to cross him. The train is moving with uniform acceleration. Find
(i) velocity of the train when the front of the first compartment crosses the observer
(ii) acceleration of the train
(Length of each compartment is 15m)

Answer 1

Hint We need to first find the velocity by dividing the length of the first compartment by the time taken for the first compartment to cross the observer. The velocity for the second compartment will be the length of the compartment divided by the time. The acceleration will be the difference in velocity by the difference in time.

Formula Used: In this solution we will be using the following formula,
$\Rightarrow v = \dfrac{l}{t}$
where $v$ is the velocity, $l$ is the length and $t$ is the time.
and $a = \dfrac{{{v_2} - {v_1}}}{{\Delta t}}$ where $a$ is the acceleration.

Complete step by step answer
In the question it is provided that the length of a compartment is $l = 15m$ and it is given that the first compartment of the train takes 2 second to cross the observer. Hence the speed of the train when it crosses the first compartment will be given by the length of the train by the time taken to cross. So we get the initial velocity of the train as,
$\Rightarrow {v_1} = \dfrac{l}{{{t_1}}}$ . Now substituting the $l = 15m$ and ${t_1} = 2s$ we get,
$\Rightarrow {v_1} = \dfrac{{15}}{2}$
On calculating we get, ${v_1} = 7.5m/s$
It is given that the second compartment of the train crosses the observer at the time given as $2.5s$ . So to get the final speed of the train we divide the length of the compartment by the time taken for the second compartment to cross the observer. Hence we get,
${v_2} = \dfrac{l}{{{t_2}}}$ . Now substituting the values $l = 15m$ and ${t_2} = 2.5s$
$\Rightarrow {v_2} = \dfrac{{15}}{{2.5}}$
On calculating we get,
$\Rightarrow {v_2} = 6m/s$
So the acceleration will be given by the change in velocity by the change in time. So we have,
$\Rightarrow a = \dfrac{{{v_2} - {v_1}}}{{\Delta t}}$
The change in time is, $\Delta t = {t_2} - {t_1}$
Substituting we get,
$\Rightarrow \Delta t = 2.5 - 2 = 0.5s$
Now substituting the values in the formula for acceleration, we have
$\Rightarrow a = \dfrac{{6 - 7.5}}{{0.5}}$
Hence we get, $a = - 3m/{s^2}$
So the velocity of the train when the front of the first compartment crosses the observer is $7.5m/s$ .
And the acceleration of the train is $- 3m/{s^2}$ .

Note
The acceleration of a body is the rate of change of the velocity of the body with respect to the time. In the answer we have the acceleration as negative. The negative acceleration means that the body is decelerating. That is the body is slowing down.