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

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

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