Question

A man walks up a stationary escalator in \[90\sec \] when this man stands on a moving escalator he goes up in \[60\] Seconds the time taken by the man to walk out the moving escalator is
\[\begin{align}
  & A.30s \\
 & B.45s \\
 & C.36s \\
 & D.48s \\
\end{align}\]

Answer 1

Hint: Relative velocity is the basic concept in order to solve this question. The speed is given as the distance per unit time which is a scalar quantity. That means it is having only magnitude and not depending on the direction. These all may help us to solve this question.

Complete answer:
It is given in the question that,
For a stationary escalator,
Time taken by the man to walks up will be given as,
$t=90s$
And also the time taken by the man to walks up in a moving escalator will be written as,
$t=60s$
Let us take the distance as $x$,
As we all know,
The speed is given as the speed divided by the time taken which can be expressed by the equation,
$v=\dfrac{d}{t}$
Where $d$ the distance taken to travel, $t$ is the time taken.
Let us substitute the values in the equations,
For the stationary escalator,
${{v}_{1}}=\dfrac{x}{90}$
For the moving escalator,
${{v}_{2}}=\dfrac{x}{60}$
Therefore the relative velocity is given as,
$v={{v}_{1}}+{{v}_{2}}$
Substituting the values in the equation,
$v=\dfrac{x}{90}+\dfrac{x}{60}$

Therefore let us simplify this equation in order to find the time taken,
$v=\dfrac{5x}{80}$
That is,
$v=\dfrac{x}{36}$
Therefore the time taken will be found by comparing with the general formula,
$\therefore t=36s$

So, the correct answer is “Option C”.

Note:
Relative motion is defined as the mathematical comparison of two or more bodies which are in motion. Acceleration is the variation in a body’s velocity in a particular period of time. Acceleration is commonly related to velocity but it is an independent variable also.