Question

Assume that there are two trains A and B of length $400m$. Each of them are in motion on two parallel tracks with a uniform speed of $72km{{h}^{-1}}$ in the identical direction, with A ahead of B. The driver of B plans to overtake A and accelerates by $1m{{s}^{-2}}$. When after $50s$, the guard of B just brushes past the driver of A, find out the original distance between them?

Answer 1

Hint: First of all convert the velocities in terms of metre per second. Then find out the distance traversed by train A by using the newton’s third equation of motion. And find the distance traversed by the train B by taking the product of the velocity and the time taken. Find out the difference between these distances. This all will help you in solving this question.

Complete step by step answer:
It has been mentioned in the question that the speed of the trains are given as,
${{u}_{A}}={{u}_{B}}=72km{{h}^{-1}}$
We have to convert this into the metre per second. That is,
${{u}_{A}}={{u}_{B}}=72\times \dfrac{5}{18}=20m{{s}^{-1}}$
By using the equation of motion,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
For the train B we can write that,
${{s}_{B}}={{u}_{B}}t+\dfrac{1}{2}a{{t}^{2}}$
The time taken has been mentioned as,
$t=50s$
Acceleration of this train is mentioned as,
$a=1m{{s}^{-2}}$
Substituting this values in the equation will give,
\[\begin{align}
  & {{s}_{B}}=20\times 50+\dfrac{1}{2}\times 1\times {{\left( 50 \right)}^{2}} \\
 & \Rightarrow {{s}_{B}}=1000+1250=2250m \\
\end{align}\]
The distance covered by the train A can be shown as,
\[{{s}_{A}}={{u}_{A}}\times t\]
Substituting the values in it will give,
\[{{s}_{A}}=20\times 50=1000m\]
Hence the original distance between the trains is given as,
\[d={{s}_{B}}-{{s}_{A}}\]
Substituting the values in it will give,
\[d=2250-1000=1250m\]
Therefore the original distance between the trains is obtained as \[1250m\].

Note: The velocity is the time rate of variation of the displacement. Distance covered is the total length of the path traversed by a body. The displacement is the shortest distance between the initial and final points. Time rate of variation of distance is known as speed. Acceleration is the time rate of variation of velocity.