Question

A person is traveling from the ground floor to the first floor in a mall using an escalator. Consider the following three separate cases. When the person is standing on the moving escalator it takes one minute for him to reach the first floor. If the escalator does not move it takes him 3 minutes to walk up the stationary escalator to reach the top. How long will it take for the person to reach the top if he walks up the moving escalator?
(A) $30\sec$
(B) $45\sec$
(C) $40\sec$
(D) $35\sec$

Answer 1

Hint: In the question, the time taken by the person to reach the top is given in both cases, when the escalator is stationary with respect to the ground and when the person is stationary with respect to the escalator. The time taken when both escalator and person move at the same time can be calculated by adding their velocities.

Complete Step by step solution:
Let the-
Distance covered by the man be s.
The velocity of the person with respect to the escalator be $v_{pe}$
The velocity of the person with respect to the ground be $v_{pg}$
The velocity of the escalator with respect to the ground be $v_{eg}$
Then the velocity of the person with respect to the escalator can be given by,
$v_{pe}=v_{pg}-v_{eg}$
In the case where the escalator is moving and the person is standing, the velocity
$v_{pe}$ becomes zero because there is no relative motion between the person and the escalator.
Here, $v_{pg}=v_{eg}$
The time taken is $1min$, writing it in seconds we have $t_1=60\sec$
Then, the velocity of the escalator with respect to ground is given by-
$v_{eg}={\displaystyle \frac{s}{60}}$
(As velocity is given by displacement over time taken)
In the second case, when the escalator is stopped and the person walks up, the velocity of the escalator with respect to the ground becomes zero.
We know that$v_{eg}=0$.
Therefore we have-
$v_{pg}=v_{pe}$
Given that the time taken to reach the top this time is $3min$or
 $t_2=3\times 60=180\sec$
The velocity of the person with respect to the escalator is,
$v_{pe}={\displaystyle \frac{s}{180}}$
In the third case, when both the escalator and the person move with their own velocities, they can be represented by-
$v_{pe}=v_{pg}-v_{eg}$
They can be written as-
${\displaystyle \frac{s}{t_2}}={\displaystyle \frac{s}{t_3}}-{\displaystyle \frac{s}{t_1}}$
$\Rightarrow {\displaystyle \frac{1}{t_3}}={\displaystyle \frac{1}{t_1}}+{\displaystyle \frac{1}{t_2}}$
On putting the values of $t_1$and $t_2$in the equation,
${\displaystyle \frac{1}{t_3}}={\displaystyle \frac{1}{60}}+{\displaystyle \frac{1}{180}}$
${\displaystyle \frac{1}{t_3}}={\displaystyle \frac{3+1}{180}}={\displaystyle \frac{4}{180}}={\displaystyle \frac{1}{45}}$
Therefore we can say that, $t_3=45\sec$
Thus it takes the person $45$ seconds to reach the top when both the escalator and the person move together.

Option (B) is correct.

Note: To represent relative velocity, we write two subscript letters below $v$, where the first letter defines the object’s velocity while the second represents the object with respect to whom this velocity is defined. For example, $v_{AB}$ will be pronounced as velocity of $A$ with respect to $B$ . It is defined as $v_{AB}=v_A-v_B$ , where $v_A$ and $v_B$ are the absolute velocities.