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?
- 36s
- 30s
- 60s
- 26s
The Correct Option is A
Solution and Explanation
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
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Concepts Used:
Motion in a straight line
The motion in a straight line is an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. It is nothing but linear motion.
Types of Linear Motion:
Linear motion is also known as the Rectilinear Motion which are of two types:
- Uniform linear motion with constant velocity or zero acceleration: If a body travels in a straight line by covering an equal amount of distance in an equal interval of time then it is said to have uniform motion.
- Non-Uniform linear motion with variable velocity or non-zero acceleration: Not like the uniform acceleration, the body is said to have a non-uniform motion when the velocity of a body changes by unequal amounts in equal intervals of time. The rate of change of its velocity changes at different points of time during its movement.