Question : A man runs at a speed of 4ms β1 to overtake a standing bus. When he is 6 m behind the door at π‘ = 0, the bus moves forward and continues with a constant acceleration of 1.2ms β2 . The man reaches the door in time π‘. Then,
(A) 4t=6+0.6t^{2}
(B) 1.2t^{2}=4t
(C) 4t^{2}=1.2t
(D) 6+4t=0.2t^{2}
Answer : option (A)
Solution :
