Question

A train starts from rest and moves with a constant acceleration of $2\,\,{\text{m/}}{{\text{s}}^{\text{2}}}$ for half a minute. The brakes are then applied and the train comes to rest in one minute. Find the total distance moved by the train.

Answer 1

Since, the train starts from rest, the initial velocity will be zero. From there, obtain the distance covered by the train for the first $30\,\,\sec$ . For the next $30\,\,\sec$ , calculate the velocity, acceleration (retardation) and substitute the value in the distance formula to obtain the distance covered by the train in the next $30\,\,\sec$ . After that calculate the total distance covered.

Complete step by step answer:
Given,
The initial velocity will be zero as the train starts from the rest.
$\therefore u = 0$
For first $30\,\,\sec$, it moves with the acceleration of $a = 2\,\,{\text{m/}}{{\text{s}}^{\text{2}}}$
Therefore, in this time interval, the distance covered is given by:
$S = ut + \dfrac{1}{2}a{t^2} \\\ \Rightarrow S = 0 + \dfrac{1}{2} \times 2 \times 30 \times 30 \\\ \Rightarrow S = 900\,\,{\text{m}}$

After 30 seconds, velocity obtained by this acceleration is given by,
$v = u + at \\\ \Rightarrow v = 0 + 2 \times 30 \\\ \Rightarrow v = 60\,\,{\text{m/s}}$

From this initial velocity, brakes are applied and after that the train comes to a stop in $60{\text{ sec}}$ .
Therefore, retardation is obtained by:
$v = u - at$
Substitute $v = 0$, $u = 60$ and $t = 60$ in the above equation.

Therefore,
\Rightarrow 0 = 60 - a \times 60 \\\ \Rightarrow a = \dfrac{{60}}{{60}} \\\ \Rightarrow a = 1\,\,{\text{m/}}{{\text{s}}^{\text{2}}} \\\
As it is retardation, therefore $a = - 1\,\,{\text{m/}}{{\text{s}}^{\text{2}}}$
Now, the distance covered by the train in this time is given by,
${v^2} = {u^2} + 2aS$
Substitute $v = 0$, $u = 60$ and $a = - 1$ in the above equation.
Therefore,
$\Rightarrow {0^2} = {\left( {60} \right)^2} + 2\left( { - 1} \right)S \\\ \Rightarrow 0 = 3600 - 2S \\\ \Rightarrow 2S = 3600 \\\ \Rightarrow S = 1800\,\,{\text{m}}$

Now, the total distance covered by the train is:
$\Rightarrow 900 + 1800\,\,{\text{m}} \\\ \therefore {\text{2700 m}}$

Hence, the required distance is ${\text{2700 m}}$.

Note: While solving this problem, it is important to remember that the train does not make a sudden halt, after the brakes are pulled; it moves a certain distance until it completely stops. It is important to take a note of the sign of the acceleration, it is negative because it acts in the direction opposite to the motion of the train.