Question

A person walked up a stalled escalator in $90\, s$ . When standing on the same escalator, now moving, he is carried up in $60\, s$. How much time would it take him to walk up the moving escalator?

• A. 36s
• B. 30s
• C. 60s
• D. 26s

Answer 1

Answer: (A)

36s

Answer 2

Let the length of escalator be $L .$
If $v$ is the velocity of man (relative to escalator) and $v$ ' of escalator.
Then according to given problem
$\frac{L}{v}=90 \,s\,\,\,...(1)$
and
$\frac{L}{\nu^{'}}=60\, s\,\,\,...(2)$
Now if the person walks up on the moving escalator his velocity relative to the ground will be $v+v^{'}$.
So, time taken by him to move a distance $L$ relative to the ground will be :
$t=\frac{L}{v+v^{'}}$
$\Rightarrow \frac{1}{t}=\frac{v^{'}}{L}+\frac{v}{L}$
which in the light of Eqs. (1) and (2), gives
$\frac{1}{t}=\frac{1}{60}+\frac{1}{90}$
$\text { i.e., } t=36\, s$

So, the correct option is (A): 36s