A bus starts moving with acceleration 2,m / s ^2 . A cyclist 96,m behind bus starts simultaneously t

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2.Motion in Straight Line

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A bus starts moving with acceleration $2\,m / s ^2$. A cyclist $96\,m$ behind the bus starts simultaneously towards the bus at $20\,m / s$. After $........\,s$ time will he be able to overtake the bus.

A

$4$

B

$8$

C

$12$

D

$16$

Solution

(b)

Let after a time t, the cyclist overtake the bus. Then $96+\frac{1}{2} \times 2 \times t^2=20 \times t$ or $t^2-20 t+$ $96=0$

$\therefore t =\frac{20 \pm \sqrt{400-4 \times 96}}{2 \times 1}$

$=\frac{20 \pm 4}{2}=8\,sec . \text { and } 12\,sec .$

Standard 11

Physics

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