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

Hint: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.