A man runs at a speed of 40 ms to overtake a standing bus He

A man runs at a speed of 4.0 m/s to overtake a standing bus. He is 6.0 meters behind the door at t=0. THe bus moves forward and continues with a constant acceleration of 1.2 m/s². How long does it take for the man to catch the door. If he starts out actually 10.0 meters from the door, will he ever catch up running at the same speed?

Solution

Srelative = Urelative + 1/2*arelative*t^2

=> 6 = 4*t + 1/2*1.2*t^2

=> t = 1.26 sec    ( time to catch bus)

at X = 10 , t = 1.937 sec ( time to catch bus) also as it take more time so he can\'t reach there with same speed