Question

Please solve it?

a child run to and fro with a speed 9km/h(with respect to the belt) oh a long horizontally moving belt between his father and mother located 50m apart ok the moving belt. the belt moves with a speed of 4km/h. for an observer on stationary platform outside what is
1.speed of the child while running in the direction of the motion of the belt?
2.speed of the child while running opposite to the direction of motion of the belt?
3.time taken by the child in (1) and (2) ?

Answer 1

Given
Speed of belt `= 4\ kmh^-1`
Let the direction of speed of belt be `hati`
Speed of child with respect to belt `v_(cb) = 9\ kmh^-1`

1. When child runs in the direction of motion of the belt his velocity is seen by a stationary observer is `v_c = 9+4=13hati\ kmh^-1`
2. When child runs in the direction opposite to that of the belt his velocity is seen by the same observer as
   `v_c=-9+4= -5hati\ kmh^-1`
3. Distance between father and mother located on the same moving belt `=50\ m`
   Velocity of child observed by the stationary observer:
   in one case `+9\ kmh^-1` and in the other case as `-9\ kmh^-1`. same as the child and parents are present on the same belt. Only difference is the direction of motion of child, i.e.
   Hence, time taken by child in both cases 1 and 2 is

   `t= "distance"/"speed"`
   Inserting given values we get
   `t= 50/((9xx1000)/3600)`
   `=>t= 20\ s`