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On a long horizontally moving belt (Fig.), a child runs to and from with a speed 9 kmh-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 kmh-1. For an observer on a stationary platform outside, what is the1. speed of the child running in the direction of motion of the belt?.2. speed of the child running opposite to the direction of motion of the belt?3. time taken by the child in (1) and (2)?Which of the answers alter If motion is viewed by one of the parents? |
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Answer» 1. Speed of the child running in the direction of motion of the belt = Speed of the child with respect to the belt + Speed of the belt with respect to the ground = 9 + 4 = 13 kmh-1. 2. Speed of the child running opposite to the direction of motion of the belt = speed of the child with respect to the belt speed of the belt with respect to the ground = 9 – 4 = 5 kmh-1. 3. Time taken by the child in both cases is the same because the speed of the child with respect to the belt does not change. Speed v = 9 kmh-1 = 2.5 ms-1 Distance d = 50 m Time t = Distance/Speed = 50/2.5 = 20 s The answer to (a) and (b) when the motion is viewed by one of the parents is 9 kmh-1. This is because the parents are also on the belt and are moving with respect to the ground. They only see the motion of the child with respect to the belt. In (c), the answer remains unaltered because the speed of the child with respect to the belt does not depend on the speed of the parents. |
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