A student is at a distance of 16 m from a bu | vedclass

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(B) Let $v$ be the minimum velocity of the student so that they can catch the bus.If the student catches the bus in time $t$,then the distance travell

Vedclass · Accepted Answer

$12$

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(B) Let $v$ be the minimum velocity of the student so that they can catch the bus.If the student catches the bus in time $t$,then the distance travelled by the student in time $t$ is equal to $16 \ m$ plus the distance travelled by the bus in time $t$.Using the equations of motion:Distance travelled by student $= v t$Distance travelled by bus $= u t + \frac{1}{2} a t^2 = 0 	imes t + \frac{1}{2} 	imes 9 	imes t^2 = 4.5 t^2$Equating the distances:$v t = 16 + 4.5 t^2$$4.5 t^2 - v t + 16 = 0$Multiply by $2$ to simplify:$9 t^2 - 2 v t + 32 = 0$For the student to catch the bus,the time $t$ must be a real value. Therefore,the discriminant of this quadratic equation must be greater than or equal to zero $(D \geq 0)$:$(-2 v)^2 - 4 	imes 9 	imes 32 \geq 0$$4 v^2 - 1152 \geq 0$$v^2 \geq 288$$v \geq \sqrt{288} = \sqrt{144 	imes 2} = 12 \sqrt{2} \ m \ s^{-1}$The minimum velocity is $12 \sqrt{2} \ m \ s^{-1}$.Comparing this with $\alpha \sqrt{2} \ m \ s^{-1}$,we get $\alpha = 12$.

$A$ student is at a distance of $16 \ m$ from a bus when the bus begins to move with a constant acceleration of $9 \ m \ s^{-2}$. The minimum velocity with which the student should run towards the bus so as to catch it is $\alpha \sqrt{2} \ m \ s^{-1}$. The value of $\alpha$ is

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