Preeti reached the metro station and found that the escalator was not working. She walked up the stationaryescalator in time t1. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t2. The time taken by her to walk up on the moving escalator will be

  1. t1 − t2
  2. \(\frac{t_{1}+t_{2}}{2}\)
  3. \(\frac{t_{1} t_{2}}{t_{2}−t_{1}}\)
  4. \(\frac{t_{1} t_{2}}{t_{2}+t_{1}}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{t_{1} t_{2}}{t_{2}+t_{1}}\)
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Detailed Solution

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Concept:

Velocity is defined as the rate of change of displacement.

\(Velocity=\frac{Displacement}{Time}\)

When two bodies move in opposite direction then their relative speed is sum of their speeds whereas if they are moving in same direction then their relative speed is difference of their speeds.

Calculation:

Velocity of girl w.r.t. elevator = \(\rm\frac{d}{t_1}=v_{ge}\)

Velocity of elevator w.r.t. ground veG \(\rm\frac{d}{t_2}\) then

velocity of girl w.r.t. ground

\(\rm\vec{v}_{gG} \ = \ \vec{v}_{ge} + \vec{v}_{eG}\)

i.e, VgG = Vge + VeG

i.e, VgG = Vge + VeG

\(\frac{d}{t}=\frac{d}{t_1}+\frac{d}{t_2}\)

\(\frac{1}{t}=\frac{1}{t_1}+\frac{1}{t_2}\)

\(t=\frac{t_1t_2}{(t_1 \ + \ t_2)} \)

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