Q. A student is standing at a distance of 50 m 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) $10\ {ms}^{- 1}$
Explanation
(d) The student is able to catch the bus if in time $t$ the distance travelled by him is equal to the distance travelled by bus in time $t$
$ie,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s_{1} = s_{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
From Eq. (i)
$\ \ \ \ \ \ \ \ 0 + \frac{1}{2}\ at^{2} = ut - d$
Or $at^{2} - 2ut + 2d = 0$
It is quadratic equation
So, $t = \frac{+ 2u \pm \sqrt{4u^{2} - 8ad}}{2} = \frac{+ 2u \pm 2\sqrt{u^{2} - 2ad}}{2}$
For $t$ to be real
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u\ \geq \sqrt{2ad} \geq \sqrt{2 \times 1 \times 50} = 10\ {ms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.