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Motion in A Straight Line question

2017 · Q145
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Motion in A Straight Line question

2017 · Q145

NEETPhysicsMotion in A Straight LineMCQ+4 / −1
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
  1. A
    t1t2t2−t1{{{t_1}{t_2}} \over {{t_2} - {t_1}}}t2​−t1​t1​t2​​
  2. B
    t1t2t2+t1{{{t_1}{t_2}} \over {{t_2} + {t_1}}}t2​+t1​t1​t2​​
  3. C
    t1−t2{{t_1} - {t_2}}t1​−t2​
  4. D
    t1+t22{{{t_1} + {t_2}} \over 2}2t1​+t2​​
View written solutionFree

Correct answer: B

Velocity of preeti with respect to elevator v1 = dt1{d \over {{t_1}}}t1​d​

Velocity of elevator with respect to ground v2 = dt2{d \over {{t_2}}}t2​d​

∴\therefore∴ Net velocity of preeti on moving escalator with respect to the ground

v = v1 + v2

dt{d \over {{t}}}td​ = dt1{d \over {{t_1}}}t1​d​ + dt2{d \over {{t_2}}}t2​d​

1t{1 \over {{t}}}t1​ = 1t1{1 \over {{t_1}}}t1​1​ + 1t2{1 \over {{t_2}}}t2​1​

∴\therefore∴ t = t1t2t1+t2{{{t_1}{t_2}} \over {{t_1} + {t_2}}}t1​+t2​t1​t2​​

Here t is the time taken by preeti to walk up on the moving escalator.

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