In the question number 97, the speed of the child running opposite to the direction of motion of the belt is

In the question number 97, the speed of the child running opposite to

1.	In the question number 97, the speed of the child running opposite to the direction of motion of the belt is
Answer» `4 KM h^(-1)` `5 km h^(-1)` `9 km h^(-1)` `13 km h^(-1)` Solution :In this case, `v_(BG) = 4 km h^(-1), v_(CB) = -9 km h^(-1)` `therefore` For an observer on a stationary platform, speed of child running opposite to the DIRECTION of motion of the belt is `v_(CG) = v_(CB) + v_(BG) = -9 km h^(-1) + 4 km h^(-1) = -5 km h^(_1)` -ve sign SHOWS that the child appears to be MOVING from right to left with a speed of `5 km h^(-1)`.