Q. A student is standing at a distance of $50\ metres$ from the bus. As soon as the bus begins its motion with an acceleration of $1\ ms^{- 2}$, 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) $52\ ms^{- 1}$
B) $8\ ms^{- 1}$
C) $10\ ms^{- 1}$
D) $12\ ms^{- 1}$
Explanation
(c) Let student catch the bus after $t$ sec. So it will cover distance $ut$
Similarly distance travelled by the bus will be $\frac{1}{2}at^{2}$ for the given condition
$ut = 50 + \frac{1}{2}at^{2} = 50 + \frac{t^{2}}{2}\ \ \ \ \ \ \lbrack a = 1\ m/s^{2}\rbrack$
$\Rightarrow u = \frac{50}{t} + \frac{t}{2}$
To find the minimum value of $u$
$\frac{du}{dt} = 0$, so we get $t = 10\ sec$, then $u = 10\ m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.