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NCERT Exemplar · Q6

Q.At a metro station, a girl walks up a stationary escalator in time t1t_1. If she remains stationary on the escalator, then the escalator take her up in time t2t_2. The time taken by her to walk up on the moving escalator will be

(a) (t1+t2)/2(t_1 + t_2)/2
(b) t1t2/(t2−t1)t_1 t_2/(t_2 - t_1)
(c) t1t2/(t2+t1)t_1 t_2/(t_2 + t_1)
(d) t1−t2t_1 - t_2
Himachal HpboseMCQ· 1mImportance★★★★★est
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When the girl walks on the moving escalator, her speed and the escalator's speed add; since speed is distance over time, the combined rate is the sum of individual rates, giving time t1t2t1+t2\boxed{\frac{t_1 t_2}{t_1 + t_2}}.

The key insight is that speeds add when motions are in the same direction. When you walk on a moving walkway, you cover ground faster than either you or the walkway alone would manage. The natural language here is rates: if the girl climbs at a certain rate (escalator-lengths per unit time) and the escalator moves at its own rate, the combined rate is simply the sum.

Let the length of the escalator be LL.

When the girl walks up the stationary escalator in time t1t_1, her walking speed is

vgirl=Lt1.v_{\text{girl}} = \frac{L}{t_1}.

When she stands still and the escalator carries her up in time t2t_2, the escalator's speed is

vesc=Lt2.v_{\text{esc}} = \frac{L}{t_2}.

Now consider what happens when both move together.

  1. The effective speed is the sum of the two speeds. The girl walks at vgirlv_{\text{girl}} relative to the escalator, and the escalator itself moves at vescv_{\text{esc}} relative to the ground. Relative to the ground, her speed is

vtotal=vgirl+vesc=Lt1+Lt2.v_{\text{total}} = v_{\text{girl}} + v_{\text{esc}} = \frac{L}{t_1} + \frac{L}{t_2}.

  1. Factor out the common length. …

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