Question

On a long horizontally moving belt (Fig.), a child runs to and fro with a speed $9km{h}^{-1}$ (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 $4km{h}^{-1}$. 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)

Answer 1

Speed of the belt, $v_B=4km/h$

Speed of the boy, $v_b=9km/h$

(a)

Since the boy is running in the same direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:

$v_{bB}=v_b+v_B=9+4=13km/h$

(b)

Since the boy is running in the direction opposite to the direction of the motion of the belt, his speed (as observed by the stationary observer) can be obtained as:

$v_{bB}=v_b-v_B=9-4=5km/h$

(c)

Distance between the child's parents = 50 m
As both parents are standing on the moving belt, the speed of the child in either direction as observed by the parents will remain the same i.e., 9 km/h = 2.5 m/s.
Hence, the time taken by the child to move towards one of his parents is 50 / 2.5 = 20 s