Question

Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t1. On the other days, if she remains stationaty 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
1) (t1+t2)/2
2) t1t2/(t2-t1)
3) t1t2/(t2+t1)
4) t1-t2

Answer 1

Hey mate here is ur ans
ur ans is option 3
see the pic for ur ans

Answer 2

Given that,

She walked up the stationary escalator in time = t₁₂

She remains stationary on the moving escalator then the escalator takes her up in time = t₂

Let the length of escalator is l.

When she walked up the stationary escalator then

The velocity of preeti w.r.to escalator will be

$v_{pe}=\dfrac{l}{t_{1}}$

When she remains stationary on the moving escalator

The velocity of escalator w.r.to ground will be

$v_{eg}=\dfrac{l}{t_{2}}$

When she to walk up on the moving escalator then,

We need to calculate the velocity of preet w.r.to ground

Using formula of relative velocity

$v_{pe}=v_{pg}-v_{eg}$

$v_{pg}=v_{pe}+v_{eg}$

Put the value into the formula

$v_{pg}=\dfrac{l}{t_{1}}+\dfrac{l}{t_{2}}$....(I)

We need to calculate the time

Using equation (I) and formula of velocity

$\dfrac{l}{t}=\dfrac{l}{t_{1}}+\dfrac{l}{t_{2}}$

$\dfrac{1}{t}=\dfrac{(t_{2}+t_{1})}{t_{1}t_{2}}$

$t=\dfrac{t_{1}t_{2}}{(t_{1}+t_{2})}$

Hence, The time taken by her to walk up on the moving escalator will be $\dfrac{t_{1}t_{2}}{(t_{1}+t_{2})}$

(3) option is correct.