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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
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Verified Answer
The correct answer is:
$\frac{t_1 t_2}{t_2+t_1}$
$\frac{t_1 t_2}{t_2+t_1}$
Let us consider, displacement is $L$, then velocity of girl with respect to ground,
$$
v_g=\frac{L}{t_1}
$$
Velocity of escalator with respect to ground,
$$
v_e=\frac{L}{t_2}
$$
Net velocity of the girl on moving escalator with respect to ground
$$
=v_g+v_e=\frac{L}{t_1}+\frac{L}{t_2} \Rightarrow v_{g e}=L\left[\frac{t_1+t_2}{t_1 t_2}\right]
$$
Now, if $t$ is total time taken by girl on moving escalator in covering distance $L$, then
$$
t=\frac{\text { distance }}{\text { speed }}=\frac{L}{L\left(\frac{t_1+t_2}{t_1 t_2}\right)}=\frac{t_1 t_2}{t_1+t_2}
$$
$$
v_g=\frac{L}{t_1}
$$
Velocity of escalator with respect to ground,
$$
v_e=\frac{L}{t_2}
$$
Net velocity of the girl on moving escalator with respect to ground
$$
=v_g+v_e=\frac{L}{t_1}+\frac{L}{t_2} \Rightarrow v_{g e}=L\left[\frac{t_1+t_2}{t_1 t_2}\right]
$$
Now, if $t$ is total time taken by girl on moving escalator in covering distance $L$, then
$$
t=\frac{\text { distance }}{\text { speed }}=\frac{L}{L\left(\frac{t_1+t_2}{t_1 t_2}\right)}=\frac{t_1 t_2}{t_1+t_2}
$$
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