Question

A person walks up an escalator at rest in a time of 60 s. When standing on the same escalator now moving with a uniform speed, he is carried up in a time of 30 s. The time taken for him to walk up the moving escalator is

• A. 10 s
• B. 15 s
• C. 20 s
• D. 25 s

Hint:

When objects move on moving platforms (like escalators), use relative motion concepts. If walking and platform speeds add up, effective speed becomes their sum. Use basic formula: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \).

Answer 1

The Correct Option is C

Step 1: Define the variables.
Let the length of the escalator be \( L \).
Let the speed of the person be \( v_p \), and speed of the escalator be \( v_e \).
From the question: \begin{itemize} \item Walking time on stationary escalator: \( \frac{L}{v_p} = 60 \) s \item Standing time on moving escalator: \( \frac{L}{v_e} = 30 \) s \end{itemize} Step 2: Calculate the speeds.
From above: \[ v_p = \frac{L}{60}, \quad v_e = \frac{L}{30} \] Step 3: Add the speeds when person walks on moving escalator.
Effective speed = \( v_p + v_e = \frac{L}{60} + \frac{L}{30} = \frac{L}{60} + \frac{2L}{60} = \frac{3L}{60} = \frac{L}{20} \) Step 4: Calculate time to walk up moving escalator.
\[ t = \frac{L}{v_p + v_e} = \frac{L}{L/20} = 20 \text{ s} \] Step 5: Select the correct option.
The calculated time is 20 s, which matches option (3).