Question

On a long horizontally moving belt, a child runs to and from 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, time taken by the child to go from father to mother and back to father is

• A. 10 s
• B. 20 s
• C. 30 s
• D. 40 s

Answer 1

Answer: (D)

40 s

Answer 2

Distance between the parents = 50m

Since the parent and child are located on the same belt, the speed of the child as observed by the stationary observer in either directions (either form mother to father or from father to mother ) will be

$9kmh^{- 1}\left( = \frac{5}{2}ms^{- 1} \right).$

$\therefore$Time taken by the child to go from father to mother and back to father is

$t = \frac{50}{(5/2)} + \frac{50}{(5/2)} = 20s + 20s = 40s.$