Question

Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t₁. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t₂. The time taken by her to walk up on the moving escalator will be

• A. \(\frac {t_1+t_2}{2}\)
• B. \(\frac {t_1t_2}{t_2-t_1}\)
• C. \(\frac {t_1t_2}{t_2+t_1}\)
• D. \(t_1-t_2\)

Answer 1

The Correct Option is C

The problem involves Preeti walking up an escalator with different scenarios: the escalator being stationary and Preeti walking on it, and the escalator moving while Preeti remains stationary. We need to find the time taken when both Preeti and the escalator are moving.

Let's break down the problem:

1. Let the speed of Preeti be \( v_p \) (when she walks up the stationary escalator).
2. Let the speed of the escalator be \( v_e \) (when Preeti is stationary on the moving escalator).
3. The total height or distance of the escalator is \( d \).

According to the problem:

• t_1 = \frac{d}{v_p} (Preeti walking on a stationary escalator)
• t_2 = \frac{d}{v_e} (Preeti stationary on a moving escalator)

We need to find \(\text{time } t\) when both are moving, i.e., when Preeti walks on the moving escalator.

• Combined speed when Preeti and the escalator both are moving: v_p + v_e
• Thus, time taken \( t \) is: t = \frac{d}{v_p + v_e}

We can express \( v_p \) and \( v_e \) in terms of \( t_1 \) and \( t_2 \) as:

• v_p = \frac{d}{t_1}
• v_e = \frac{d}{t_2}

Substituting these into the formula for \( t \):

• t = \frac{d}{\frac{d}{t_1} + \frac{d}{t_2}}
• Simplifying this gives: t = \frac{t_1t_2}{t_1 + t_2}

Therefore, the time taken by Preeti to walk up on the moving escalator is \(\frac{t_1t_2}{t_1 + t_2}\).