Question

At a metro station, a girl walks up a stationary escalator in time $t_{1}$. If she remains stationary on the escalator, then the escalator take her up in time $t_{2}$. The time taken by her to walk up on the moving escalator will be

• A. $\frac{(t_{1} + t_{2})}{2}$
• B. $\frac{t_{1}t_{2}}{(t_{2} - t_{1})}$
• C. $\frac{t_{1}t_{2}}{(t_{2} + t_{1})}$
• D. $t_{1} - t_{2}$

Answer 1

Answer: (C)

$\frac{t_{1}t_{2}}{(t_{2} + t_{1})}$

Answer 2

Let L be the length of escalator

Velocity of girl w.r.t. escalator, $v_{ge} = \frac{L}{t_{1}}$

Velocity of escalator, $v_{e} = \frac{t}{t_{2}}$

$\therefore$velocity of girl w.r.t. grounds would be

$v_{g} = v_{ge} + v_{e} = L\left( \frac{1}{t_{1}} + \frac{1}{t_{2}} \right)$

$\therefore$ The desired time is

$t = \frac{L}{v_{s}} = \frac{L}{L\left( \frac{1}{t_{1}} + \frac{1}{t_{2}} \right)} = \frac{t_{1}t_{2}}{t_{1} + t_{2}}$