Question 140: 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 t
1
. On another days, if she remains stationary  on the the moving escalator, then the escalator takes her up in time t
2
. 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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