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Question

A man is 25 m behind a bus, when the bus starts accelerating at 2 m/s2 and man starts moving with constant velocity of 10 m/s. Time taken by him to board the bus is

A
2 s
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B
3 s
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C
4 s
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D
5 s
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Solution

The correct option is D 5 s
At the time when the man boards the bus,
Displacement of man = displacement of bus + 25
i.e. 10t=12×2×t2+25
10t=t2+25
t210t+25=0
t25t5t+25=0
(t5)2=0
Solving this equation we get t=5 s

Alternate Sol:

Displacement of man w.r.t the bus, SMB=25
Initial velocity of man w.r.t the bus, uMB=10 m/s
Acceleration of man w.r.t the bus, aMB=2 m/s2
Using second equation of motion,
SMB=uMBt+12aMBt2
i.e., 25=10t12×2×t2
t210t+25=0
Solving the equation, we get t=5 sec

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