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Question

A bus of 6 m long starts moving with an acceleration of 2ms−2. A cyclist 90 m behind the bus starts simultaneously towards the bus at 20 ms−1. After what minimum 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

For boy to overtake,
In t seconds bus travels a distance of x meters and boy travels (x+90+6) meters

Using second equation of motion for bus,
x=ut+at22,
x=at22=2×t22=t2...........................(1)

For boy, Distance=speed×time
x+96=vt,
x+96=20t .........................................(2)

using (1) and (2),
t220t+96=0
we get t=12 and 8,
Since minimum time,
Hence, boy will catch the bus at t=8 seconds.


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