Question

A train is running at a speed of $40\,km{\kern 1pt} h{r^{ - 1}}$ and it crosses a post in $18\sec $. What is the length of the train?

Answer 1

Hint: In order to solve this question we need to understand the distance, displacement and velocity of particles. So distance is defined as a physical quantity which is equal to path length covered by a particle, it is scalar quantity. Displacement is defined as the shortest path length between initial and final points of a particle motion, it is a vector quantity. Speed is defined as distance covered by particle per unit time, it is a scalar quantity.Velocity is defined as displacement covered by particle per unit time, it is a vector quantity.

Complete step by step answer:
Speed of the train is given as, $v = 40\,km{\kern 1pt} h{r^{ - 1}}$. We have to convert the train speed in $m{\kern 1pt} {\kern 1pt} {\sec ^{ - 1}}$ so to do the same, speed in $m{\kern 1pt} {\kern 1pt} {\sec ^{ - 1}}$ is,
$v = 40 \times \dfrac{5}{{18}}\,m{\kern 1pt} {\kern 1pt} {\sec ^{ - 1}}$
since $1Kmh{r^{ - 1}} = \dfrac{5}{{18}}\,m{s^{ - 1}}$
$\Rightarrow v = 11.11\,m{\kern 1pt} {\sec ^{ - 1}}$
Time taken by train to cross a post is, $t = 18\sec $

Let the length of the train be “L”.So to cover the post, the whole length of the train must go past by post. So from the definition of speed we get, $v = \dfrac{d}{t}$.So the distance or length of train covered is, $d = vt$.Putting values we get,
$L = (11.11m{\sec ^{ - 1}}) \times (18\sec )$
$\therefore L = 200\,m$

Therefore the length of train is, $L = 200\,m$.

Note: It should be remembered that, here we have assumed that the motion is analyzed from a static frame of reference or inertial frame of reference in which the observer is assumed to be at rest. If the motion is analyzed from an observer which is also moving with respect to the train, then in that case we have to relatively add or subtract velocity (depending on direction) to get the final result.