A boy reaches the airport and find that the escalator is not working. He walks up the stationary escalator in time t1. If he remains stationary on a moving escalator then the escalator takes him up in time t2. The time taken by him to walk up on the moving escalator will be:

Option 1 -

t 1 t 2 t 2 + t 1

Option 2 -

t 2 t 1

Option 3 -

t 1 + t 2 2

Option 4 -

t 1 t 2 t 2 t 1

0 2 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    a month ago
    Correct Option - 1


    Detailed Solution:

    Let the distance travelled is x, so

    | v E g | = x t 2 speed of escalator for ground

    | v B E | = x t 1 speed of boy concerning the escalator

    | v B g | = x t 1 + x t 2 speed of boy concerning ground

    The time taken by him to walk up the moving escalator = t =  x | v B g |

    t = x x t 1 + x t 2 = t 1 t 2 t 1 + t 2

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alok kumar singh

Kindly go through the solution '

 

A
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V
Vishal Baghel

[h] = ML2T-1

[E] = ML2T-2

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A
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According to question, we can write

 10 = 1 2 a t 2 . . . . . . . . . . . . . . . ( 1 ) a n d                        

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X = 1 2 A [ ( 2 t ) 2 t 2 ] = 3 ( 1 2 a t 2 ) = 3 0 m  

 

A
alok kumar singh

Average speed = 4 v 2 3 v

= 4 v 3

(d) Initial velocity  = - v j ˆ

Final velocity = v i ˆ

Change in velocity  = v i ˆ - ( - v j ˆ )

= v ( i ˆ + j ˆ ) Momentum gain is along i ˆ + j ˆ

 Force experienced is along i ˆ + j ˆ

  Force experienced is in North-East direction.

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