At a metro station, a girl walks up a stationary escalator in time t1, If she remains stationary on the escalator, then the escalator take her up in time t2. The time taken by her to walk up on the moving escalator will be

Question

At a metro station, a girl walks up a stationary escalator in time t1, If she remains stationary on the escalator, then the escalator take her up in time t2. The time taken by her to walk up on the moving escalator will be (A) (t1 + t2)/2 (B) t1 t2 /(t2 — t1) (C) t1t2/(t2 + t1) (D) t1 - t2

Tardigrade Admin · Accepted Answer

We have to find net velocity with respect to the Earth that will be equal to velocity of the girl plus velocity of escalator. Let displacement is

L

, then

Velocity of girl

v_{g} = \frac{L}{t _{1}}

Velocity of escalator

v_{e} = \frac{L}{t _{2}}

Net velocity of the girl

= v_{g} + v_{e} = \frac{L}{t _{1}} + \frac{L}{t _{2}}

If

t

is total time taken in covering distance

L

, then

\frac{L}{t} + \frac{L}{t _{1}} + \frac{L}{t _{2}}

\Rightarrow t = \frac{t _{1} t _{2}}{t _{1} + t _{2}}

Q. 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

$(t_{1} + t_{2}) /2$

$t_{1} t_{2} / (t_{2} — t_{1})$

$t_{1} t_{2} / (t_{2} + t_{1})$

$t_{1} - t_{2}$

Q. At a metro station, a girl walks up a stationary escalator in time t1​, If she remains stationary on the escalator, then the escalator take her up in time t2​. The time taken by her to walk up on the moving escalator will be

(t1​+t2​)/2

t1​t2​/(t2​—t1​)

t1​t2​/(t2​+t1​)

t1​−t2​

Q. 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

$(t_{1} + t_{2}) /2$

$t_{1} t_{2} / (t_{2} — t_{1})$

$t_{1} t_{2} / (t_{2} + t_{1})$

$t_{1} - t_{2}$