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(B) Let $t$ be the time in seconds after which the cyclist overtakes the bus.For the bus,the distance covered from its starting point is $s_{bus} = u_

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$8$

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(B) Let $t$ be the time in seconds after which the cyclist overtakes the bus.For the bus,the distance covered from its starting point is $s_{bus} = u_{bus}t + \frac{1}{2}at^2 = 0 + \frac{1}{2} 	imes 2 	imes t^2 = t^2$.The cyclist starts $96\,m$ behind the bus,so the total distance the cyclist must cover to reach the bus is $s_{cyclist} = 96 + t^2$.Since the cyclist moves at a constant speed of $20\,m/s$,the distance covered by the cyclist is $s_{cyclist} = 20t$.Equating the two expressions for the distance: $96 + t^2 = 20t$.Rearranging into a quadratic equation: $t^2 - 20t + 96 = 0$.Solving for $t$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$t = \frac{20 \pm \sqrt{400 - 4 	imes 96}}{2} = \frac{20 \pm \sqrt{400 - 384}}{2} = \frac{20 \pm \sqrt{16}}{2} = \frac{20 \pm 4}{2}$.This gives two possible times: $t_1 = \frac{24}{2} = 12\,s$ and $t_2 = \frac{16}{2} = 8\,s$.The cyclist will first overtake the bus at $t = 8\,s$.

$A$ bus starts moving with an acceleration of $2\,m/s^2$. $A$ cyclist,$96\,m$ behind the bus,starts moving simultaneously towards the bus at a constant speed of $20\,m/s$. After how many seconds will the cyclist be able to overtake the bus?

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