Question

A student is standing at a distance of $50$ meters behind a bus. As soon as the bus starts with an acceleration of $1m{s}^{-2}$, the student starts running towards the bus with a uniform velocity $u$. Assuming the motion to be along straight road the minimum value of $u$, so that the student is able to catch the bus is:

• A. $5m{s}^{-1}$
• B. $8m{s}^{-1}$
• C. $10m{s}^{-1}$
• D. $12m{s}^{-1}$

Answer 1

The correct option is C $10m{s}^{-1}$

For bus $s=\frac{1}{2}a{t}^2\Rightarrow s=\frac{1}{2}\times 1\times t^2$......$\left(i\right)$

For student $(50+s)=ut$........$\left(ii\right)$

from $\left(i\right)$ and $\left(ii\right)$ $50+\frac{t^2}{2}=ut$

$\Rightarrow t^2-2ut+100=0$

$t=\frac{2u\pm \sqrt{4u^2-4\times 100}}{2}$

t must be real so $4u^2-400=0$ for $t$ to be minimum

$u=10m/s$