Question

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

• A. $5\frac{m}{s}$
• B. $8\frac{m}{s}$
• C. $10\frac{m}{s}$
• D. $12\frac{m}{s}$

Answer 1

The correct option is C $10\frac{m}{s}$

Let, the student catches the bus after $tsec$. So, it will cover distance $ut$.

Similarly, distance travelled by the bus will be $\frac{1}{2}a{t}^2$. Therefore, for the given condition
$ut=50+\frac{1}{2}a{t}^2=50+\frac{t^2}{2}\Rightarrow u=\frac{50}{t}+\frac{t}{2}(asa=1\frac{m}{s^2})....\left(1\right)$

So, to find the minimum value of u, $\frac{du}{dt}=0,\Rightarrow \frac{d(\frac{50}{t}+\frac{t}{2})}{dt}=0\Rightarrow \frac{-50}{t^2}+\frac{1}{2}=0\Rightarrow t^2=100\Rightarrow t=10s$

Then putting value of t in equation $\left(1\right)$, we get $u=10\frac{m}{s}$