Question

At Janakpuri west (magenta line) metro station, a boy walks up a stationary escalator in time \[{t_1}\] . If he remains stationary on the escalator, then the escalator takes him up in time \[{t_2}\] . The time taken by him to walk up on the moving escalator will be:
A. \[\dfrac{{{t_1} + {t_2}}}{2}\]
B. \[\dfrac{{{t_1}{t_2}}}{{{t_1} + {t_2}}}\]
C. \[{t_1} - {t_2}\]
D. \[\dfrac{{{t_1}{t_2}}}{{{t_1} - {t_2}}}\]

Answer 1

Hint: First of all, we will find the time taken by the boy to cover the distance when the escalator is stationary. After that we will find the time taken by the escalator to move up when the boy is stationary. We will again find the relative velocity of the boy with respect to the ground. Then we will find the total time taken.

Formula used:
We will use the formula which gives the velocity of an object as shown below:
\[v = \dfrac{d}{t}\] …… (1)
Where,
\[v\] indicates the velocity.
\[d\] indicates the distance.
\[t\] indicates the total time taken.

Complete step by step answer:
In the given question, we are supplied with the following data:
A boy walks up a stationary escalator in time \[{t_1}\] .We are asked to find the time taken by him to walk up on the moving escalator.

Let us proceed to solve the numerical. Let the speed of the boy when the escalator is stationary be \[{v_1}\] .The speed of the escalator itself be \[{v_2}\] .Again, we will consider the slant height or the distance be \[d\] .

First, the time taken by the boy to go up when the escalator is completely at halt is given by:
${t_1} = \dfrac{d}{{{v_1}}} \\
\Rightarrow {v_1} = \dfrac{d}{{{t_1}}}$

Secondly, the time taken to go up when the boy is completely at halt and only the escalator is moving:
${t_2} = \dfrac{d}{{{v_2}}} \\
\Rightarrow {v_2} = \dfrac{d}{{{t_2}}} \\$
As the both the boy and the escalator is moving in the same direction, the relative speed of the boy with respect to the ground is:
\[v = {v_1} + {v_2}\]

Now, we use the equation (1) and substitute the required values:
$t = \dfrac{d}{v} \\
\Rightarrow t = \dfrac{d}{{{v_1} + {v_2}}} \\
\Rightarrow t = \dfrac{d}{{\dfrac{d}{{{t_1}}} + \dfrac{d}{{{t_2}}}}} \\
\therefore t = \dfrac{{{t_1}{t_2}}}{{{t_1} + {t_2}}}$

Hence, the time taken by the boy is \[\dfrac{{{t_1}{t_2}}}{{{t_1} + {t_2}}}\]. The correct option is B.

Note: While solving this problem, one thing we must be very clear that in the first condition, the escalator was stationary and while in the second condition the boy is stationary. We should also remember that we need to find the relative velocity of the boy with respect to the ground, which is the summation of velocities, as both the velocities are in the same direction.