Question

A student is standing at a distance of $50$ metre from the bus. As soon as the bus begins its motion with an acceleration of $1{ms}^{-2}$, the students 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. $8{ms}^{-1}$
• B. $5{ms}^{-1}$
• C. $12{ms}^{-1}$
• D. $10{ms}^{-1}$

Answer 1

The correct option is D $10{ms}^{-1}$

Acceleration of the bus $a=1m{s}^{-2}$

Let the distance traveled by bus be $x$ when the student caught it after time $t$.

Velocity of the student $=u$

Initial velocity of bus $u_1=0$

$\therefore$ $x=\frac{1}{2}a{t}^2$ $⟹x=\frac{t^2}{2}$ $(\because a=1m{s}^{-2})$

Distance traveled by student $=x+50$

$\therefore$ $x+50=ut$

$\frac{t^2}{2}+50=ut$ $⟹t^2-2ut+100=0$

The above relation must have real roots i.e its discriminant $\ge 0$

$(2u{)}^2-4\times 100\ge 0$

OR $u^2-100\ge 0$ $⟹u>10m{s}^{-1}$

Thus the minimum velocity of the student must be $10m{s}^{-1}$