Question

Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time ${{t}_{1}}$. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time ${{t}_{2}}$. The time taken by her to walk up on the moving escalator will be:
$A)\text{ }{{t}_{1}}-{{t}_{2}}$
$B)\text{ }\dfrac{{{t}_{1}}+{{t}_{2}}}{2}$
$C)\text{ }\dfrac{{{t}_{1}}{{t}_{2}}}{{{t}_{2}}-{{t}_{1}}}$
$D)\text{ }\dfrac{{{t}_{1}}{{t}_{2}}}{{{t}_{2}}+{{t}_{1}}}$

Answer 1

Hint: To solve this problem, we must find Preeti’s speed of walking and the speed of the elevator individually. When she walks up the moving elevator, her total speed will be the sum of her individual speed of walking and the speed of the elevator. By using the relation between speed, distance and time, we can get the time taken in this case.

Formula used:
$\text{Speed = }\dfrac{\text{Distance}}{\text{Time}}$

Complete step by step answer:
We will solve this problem by finding out the speeds of Preeti walking, and the elevator and sum them up to get the speed when Preeti walks on the moving elevator and use this speed to find the time taken in this case.
Hence, let us analyze the question.
Let the length of the path that is taken by the elevator be $l$.
Let the speed of Preeti’s walking be ${{v}_{P}}$ and the speed of the elevator be ${{v}_{E}}$.
It is given that the time taken by Preeti to walk up the length of the stationary escalator is ${{t}_{1}}$.
Also, the time taken by the elevator to move up the length with Preeti stationary on it is ${{t}_{2}}$.
Now,
$\text{Speed = }\dfrac{\text{Distance}}{\text{Time}}$ --(1)
Therefore, using (1), we get,
${{v}_{P}}=\dfrac{l}{{{t}_{1}}}$ -(2)
${{v}_{E}}=\dfrac{l}{{{t}_{2}}}$ --(3)
As explained, when she walks up the moving escalator, her total speed $v$ will be the sum of the speed of the elevator and her individual speed.Hence,
$v={{v}_{P}}+{{v}_{E}}$
Using (2) and (3), we get,
$v=\dfrac{l}{{{t}_{1}}}+\dfrac{l}{{{t}_{2}}}=l\left( \dfrac{1}{{{t}_{1}}}+\dfrac{1}{{{t}_{2}}} \right)=l\left( \dfrac{{{t}_{2}}+{{t}_{1}}}{{{t}_{1}}{{t}_{2}}} \right)=\dfrac{l}{\left( \dfrac{{{t}_{1}}{{t}_{2}}}{{{t}_{2}}+{{t}_{1}}} \right)}$ --(4)
Now, let the total time taken in the required case be $t$.
Hence, using (1), we get,
$v=\dfrac{l}{t}$ --(5)
Therefore, equating (4) and (5), we get,
$\dfrac{l}{t}=\dfrac{l}{\left( \dfrac{{{t}_{1}}{{t}_{2}}}{{{t}_{2}}+{{t}_{1}}} \right)}$
$\therefore t=\left( \dfrac{{{t}_{1}}{{t}_{2}}}{{{t}_{2}}+{{t}_{1}}} \right)$
Hence, the required time taken is $\dfrac{{{t}_{1}}{{t}_{2}}}{{{t}_{2}}+{{t}_{1}}}$.
Therefore, the correct answer is $D)\text{ }\dfrac{{{t}_{1}}{{t}_{2}}}{{{t}_{2}}+{{t}_{1}}}$.

Note: Students must be careful of the direction in which Preeti was moving. For example if she was trying to walk down the elevator moving up, then her total speed would have been the difference of the speeds of the elevator and her walking. In this problem, we added the speeds since both Preeti and the elevator are moving in the same direction individually, that is, moving up the elevator.