3.25 On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h –1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h–1. For an observer on a stationary platform outside, what is the

(a) Speed of the child running in the direction of motion of the belt?

(b) Speed of the child running opposite to the direction of motion of the belt?

(c) Time taken by the child in (a) and (b)?

Which of the answers alter if motion is viewed by one of the parents?

94 Views|Posted 9 months ago
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1 Answer
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9 months ago
3.25 Speed of the child = 9 km/h

Speed of belt = 4 km/h

(a) When the boy runs in the direction of motion of the belt, then the speed of the child observed by the stationary observer = 9 + 4 = 13 km/h

 

(b) When the boy runs in the opposite direction of motion of the belt, then the speed of the child o

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According to question, we can write

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

1 0 + x = 1 2 a ( 2 t ) 2 . . . . . . . . . . . . . . . . ( 2 )  

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

 

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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physics ncert solutions class 11th 2023

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