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A bus starts from rest and moves with an acceleration of $1 m / s^{2}$ . A boy, who is $48 m$ behind the bus run after with a constant speed of $10 m / s$ . The boy can catch the bus

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only once, after $8 s$ form start

only once, after $12 s$ from start

twice after $8 s a n d 12 s$ from start

never

answer is C.

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Detailed Solution

$48 + \frac{1}{2} \times 1 \times t^{2} = 10 t ∴ t = 8 s and 12 s .$

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