Question

A man is 45 m behind the bus when the bus start accelerating from rest with acceleration 2.5 m/s². With what minimum velocity should the man start running to catch the bus?

• A. 12 m/s
• B. 14 m/s
• C. 15 m/s
• D. 16 m/s

Answer 1

Answer: (C)

15 m/s

Answer 2

Let man will catch the bus after 't' sec . So he will cover distance ut.

Similarly distance travelled by the bus will be $\frac{1}{2}at^{2}$. For the given condition

$u\mspace{6mu} t = 45 + \frac{1}{2}a\mspace{6mu} t^{2} = 45 + 1.25\mspace{6mu} t^{2}$ $\lbrack As\mspace{6mu} a = 2.5m/s^{2}\rbrack$

⇒ $u = \frac{45}{t} + 1.25\mspace{6mu} t$

To find the minimum value of u

$\frac{du}{dt} = 0$ so we get $t = 6\sec$then,

$u = \frac{45}{6} + 1.25 \times 6 = 7.5 + 7.5 = 15m/s$