Question 135: Motion in 1D

Question

Preeti reached the metro station  and found that the escalator was not working. She walked up  the sationary escalator in time t1. On another days, if she remains stationary  on the 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

Accepted Answer

Velocity of preeti with respect to elevator v
1
= $${d \over {{t_1}}}$$
Velocity of elevator with respect to ground v
2
= $${d \over {{t_2}}}$$
$$ 	herefore $$ Net velocity of preeti on moving escalator with respect to the ground
v = v
1
+ v
2
$${d \over {{t}}}$$ = $${d \over {{t_1}}}$$ + $${d \over {{t_2}}}$$
$${1 \over {{t}}}$$ = $${1 \over {{t_1}}}$$ + $${1 \over {{t_2}}}$$
$$ 	herefore $$ t = $${{{t_1}{t_2}} \over {{t_1} + {t_2}}}$$
Here t is the time taken by preeti to walk up on the moving escalator.

Answer

$${{{t_1}{t_2}} \over {{t_2} - {t_1}}}$$

Answer

$${{{t_1}{t_2}} \over {{t_2} + {t_1}}}$$

Answer

$${{t_1} - {t_2}}$$

Answer

$${{{t_1} + {t_2}} \over 2}$$

Motion in 1D

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