A man is 25m behind a bus, when the bus starts accelerating at 2m/s2 and man starts moving with constant velocity of 10m/s. Time taken by him to board the bus is
A
2s
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B
3s
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C
4s
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D
5s
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Solution
The correct option is D5s 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 ⇒t2−10t+25=0 ⇒t2−5t−5t+25=0 ⇒(t−5)2=0 Solving this equation we get t=5s
Alternate Sol:
Displacement of man w.r.t the bus, SMB=25 Initial velocity of man w.r.t the bus, uMB=10m/s Acceleration of man w.r.t the bus, aMB=−2m/s2 Using second equation of motion, SMB=uMBt+12aMBt2 i.e., 25=10t−12×2×t2 ⇒t2−10t+25=0 Solving the equation, we get t=5sec