A bus starts moving with acceleration 2 2. ms^-2 A cyclist 96m behind the bus stats simultaneously towards the bus at 20 m/s. After what time will he be able to overtake the bus? (Least time in sec)

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answer is 8.

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Detailed Solution

$\frac{1}{2} \times 2 \times t^{2} + 96 = 20 \times t \Rightarrow t^{2} - 20 t + 96 = 0$

This gives, t = 8 or 12sec.
Hence, the cyclist will overtake after 8sec.

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