(a) Speed of the belt, v₁ = 4 km/h
Speed of the boy w.r.t belt, v₂ = 9 km/h
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 = v₁ + v₂ = 4 + 9 = 13 km/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 = v₁ + (– v₂) = 4 – 9 = -5 km/h (negative as child is moving in opposite direction)

(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 given by

$=\frac{50}{2.5}$
(d) If the motion is viewed by any one of the parents, Solutions obtained in (a) and (b) get altered. This is because the child and his parents are standing on the same belt and hence, are equally affected by the motion of the belt. Therefore, for both parents (irrespective of the direction of motion) the speed of the child remains the same i.e., 9 km/h.
For this reason, it can be concluded that the time taken by the child to reach any one of his parents remains unaltered.