Question

Two trains one travelling at $15\,m{s^{ - 1}}$ and other at $20\,m{s^{ - 1}}$ are headed towards one another on a straight track. Both the drivers apply brakes simultaneously when they are $500\,m$ apart. If each train has a retardation of $1\,m{s^{ - 1}}$ the separation after they stop is
(A) $192.5\,m$
(B) $225.5\,m$
(C) $187.5\,m$
(D) $155.5\,m$

Answer 1

Hint:Apply the stopping distance formula for the two trains and calculate the value for the both Sum up the stopping distance of the both. Subtract this distance from the distance between the two trains at a time when the brake is applied by the driver.

Useful formula:
The formula of the stopping distance is given by

$d = \dfrac{{{v^2}}}{{2\mu g}}$

Where $d$ is the stopping distance after applying brake, $v$ is the velocity, $\mu $ is the friction coefficient and $g$ is the acceleration due to gravity.

Complete step by step solution:
The speed of the first train, ${v_1} = 15\,m{s^{ - 1}}$
The speed of the second train, ${v_2} = 20\,m{s^{ - 1}}$
The distance between them when the drivers apply brake, $D = 500\,m$
By using the formula of the stopping distance and substituting the known values in it.

$d = \dfrac{{{v^2}}}{{2\mu g}}$
$d = \dfrac{{{{15}^2}}}{{2 \times 1}}$

By performing the various arithmetic operations, we get

$d = 112.5\,m$

Similarly calculating the stopping distance for the second train.

$d = \dfrac{{{{20}^2}}}{{2 \times 1}}$
${d_2} = 200\,m$

The separation between the two trains after the stop is calculated by subtracting the distance when braking the distance that the train can travel after applying the brake.

$s = 500 - \left( {112.5 + 200} \right)$

By simplifying the right side of the equation, we get

$s = 187.5\,m$

Thus the option (C) is correct.

Note:Remember that the stopping distance is the distance that is travelled for a sometime by the vehicle even after applying brake. This depends on the frictional force between the brake rubber and the wheel and also the wheel and the road.