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Question

Physics Question on Motion in a straight line

A person walks up a stationary escalator in time t1t_1. If he remains stationary on the escalator, then it can take him up in time t2t_2. How much time would it take him to walk up the moving escalator?

A

t1+t22 \frac{t_1 + t_2}{2}

B

t1t2\sqrt{t_1 t_2}

C

t1t2t1+t2 \frac{t_1 t_2}{t_1 + t_2}

D

t1+t2 t_1 +t_2

Answer

t1t2t1+t2 \frac{t_1 t_2}{t_1 + t_2}

Explanation

Solution

Let LL be the length of escalator.
Speed of man w.r.t. escalator is vmc=Lt1v_{mc} = \frac{L}{t_1}
Speed of escalator is vc=Lt2v_c = \frac{L}{t_2}
\therefore Speed of man with respect to ground would be
vm=vmc+vc=L(1t1+1t2)v_m = v_{mc} + v_c = L \left( \frac{1}{t_1} + \frac{1}{t_2} \right)
\therefore Time taken t=LL(1t1+1t2)=t1t2t1+t2t = \frac{L}{L\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}\right)}=\frac{t_{1}t_{2}}{t_{1}+t_{2}}