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

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

Aicharya · Accepted Answer

Let $v$ be the minimum velocity of student so,that he could catch the bus

If student catch the bus in time $t$, then distance travelled by student in time $t$ $=16+$ distance travelled by bus in time $t$. $\begin{aligned} & \Rightarrow v t=16+\left(u t+\frac{1}{2} a t^2\right) \\ & \Rightarrow v t=16+0 \times t+\frac{1}{2} \times 9 \times t^2 \\ & \Rightarrow v t=16+\frac{9}{2} t^2 \Rightarrow 9 t^2-2 v t+32=0\end{aligned}$ The above equation must have real roots. i.e its discriminant $\geq 0$ i.e $(2 v)^2-4 \times 9 \times 32 \geq 0$ $\begin{aligned} & \Rightarrow 4 v^2-4 \times 288 \geq 0 \Rightarrow v^2-288 \geq 0 \\ & \Rightarrow v^2 \geq 288\end{aligned}$ $v \geq 12 \sqrt{2} \mathrm{~m} / \mathrm{s}$ Minimum velocity of student to catch the bus $=12 \sqrt{2} \mathrm{~m} / \mathrm{s}=\alpha \sqrt{2} \mathrm{~m} / \mathrm{s}$ (given) $\therefore \alpha=12$

A student is at a distance $16 \mathrm{~m}$ from a bus when the bus begins to move with a constant…

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