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 1ms–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

(1) 5 ms–1

(2) 8 ms–1

(3) 10 ms–1

(4) 12 ms–1

Concept Videos :-

#45-Relative motion
#46-Relative motion-1
#47-Relative motion-2
#48-Relative motion-3
#49-Relative motion-4
#50-Relative motion-5

Concept Questions :-

Relative motion in 1-D

(3) Let student will catch the bus after t sec. So it will cover distance ut.

Similarly distance travelled by the bus will be $\frac{1}{2}a{t}^{2}$ for the given condition

$ut=50+\frac{1}{2}a{t}^{2}=50+\frac{{t}^{2}}{2}$ [$a=1\text{\hspace{0.17em}}m/{s}^{2}$

u $=\frac{50}{t}+\frac{t}{2}$

To find the minimum value of

$\frac{du}{dt}=0$, so we get $t=10\mathrm{sec}$, then $u=10\text{\hspace{0.17em}}m/s$

Difficulty Level:

• 11%
• 16%
• 71%
• 4%