# A man runs at a speed of 4ms −1 to overtake a standing bus. When he is 6 m behind the door at 𝑡 = 0, the bus moves forward and continues with a constant acceleration of 1.2ms −2 . The man reaches the door in time 𝑡. Then,

Question : A man runs at a speed of 4ms −1 to overtake a standing bus. When he is 6 m behind the door at 𝑡 = 0, the bus moves forward and continues with a constant acceleration of 1.2ms −2 . The man reaches the door in time 𝑡. 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 :

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