Question

A bus begins to move with a=1 m/s^2 a man who is 48 metre behind the bs starts running at 10 m/s to catch the bus then men will be able to catch the bus after how many seconds?

Answer 1

by taking the bus at rest bus acceleration will be given to man in opp direction then by using S=ut+1/2at^2 we can find the value of t

Answer 2

Let assume
BUS as 'B' and man as 'A'

hence he reaches bus after 8s or 12s

here,
aAB = acceleration of A(man) with respect to B(bus)
aAG = acceleration of A with respect to the ground
aBG = acceleration of B with respect to the ground

VAB = velocity of A with respect to the bus
VAG = velocity of A with respect to the ground
VBG = velocity of B with respect to the ground