On a long horizontally moving belt,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?

(N/A) Speed of the belt,$v_{B} = 4 \; km \; h^{-1}$.
Speed of the child with respect to the belt,$v_{c} = 9 \; km \; h^{-1}$.
$(a)$ Since the child is running in the same direction as the motion of the belt,his speed as observed by the stationary observer is:
$v_{a} = v_{c} + v_{B} = 9 + 4 = 13 \; km \; h^{-1}$.
$(b)$ Since the child is running opposite to the direction of motion of the belt,his speed as observed by the stationary observer is:
$v_{b} = v_{c} - v_{B} = 9 - 4 = 5 \; km \; h^{-1}$.
$(c)$ Distance between the parents,$d = 50 \; m = 0.05 \; km$.
Since both parents are on the moving belt,the relative speed of the child with respect to the parents is always $9 \; km \; h^{-1}$.
Converting speed to $m \; s^{-1}$: $9 \; km \; h^{-1} = 9 \times \frac{5}{18} = 2.5 \; m \; s^{-1}$.
Time taken,$t = \frac{d}{v_{c}} = \frac{50 \; m}{2.5 \; m \; s^{-1}} = 20 \; s$.
If the motion is viewed by one of the parents,the answers for $(a)$ and $(b)$ will alter because the parents are in the same frame of reference as the belt. For the parents,the speed of the child is always $9 \; km \; h^{-1}$ regardless of the direction. The time taken remains unaltered.