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

The Correct Option is A

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