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Question

A bus starts moving with acceleration 2 msāˆ’2. A cyclist 96 m behind the bus starts simultaneously towards the bus at a constant speed of 20 m/s. After what time will he be able to overtake the bus?

A
4 s
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B
8 s
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C
12 s
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D
16 s
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Solution

The correct option is B 8 s
Let us assume that the cyclist overtakes the bus after time t seconds. The bus will cover a distance ‘S’ in time t seconds. Then the cyclist has to cover a distance (96+S) m in order to overtake the bus.
96 + (distance travelled by bus in time t) = (distance travelled by cyclist in time t)
12×2×t2+96=20×t
t220t+96=0
This gives, t=8 s or 12 s. Hence, the cyclist will overtake the bus at t=8 s.
After t=12 s bus will again overtake the cyclist as bus has some acceleration and cyclist moves with constant speed.

Alternate Solution:
Relative displacement, Srel=96 m
Relative velocity of cyclist w.r.t bus, urel=200=20 m/s
Relative acceleration, arel=02=2 m/s2
Using second equation of motion,
Srel=urelt+12arelt2
96=20t12×2×t2
t220t+96=0
This gives, t=8 s or 12 s
Hence, the minimum time required by the cyclist to overtake the bus 8 s

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