MCQ
A man is $45 \,m$ behind the bus when the bus start accelerating from rest with acceleration $2.5$ $m/s^2$. With what minimum velocity should the man start running to catch the bus ?.......$m/s$
- A$12$
- B$14 $
- ✓$15$
- D$16 $
✓
Answer
Correct option: C.
$15$
c
(c) Let man will catch the bus after $'t'$ sec . So he will cover distance $ut$.
(c) Let man will catch the bus after $'t'$ sec . So he will cover distance $ut$.
Similarly distance travelled by the bus will be $\frac{1}{2}a{t^2}$. For the given condition
$u\;t = 45 + \frac{1}{2}a\;{t^2}$$ = 45 + 1.25\;{t^2}$ $[As\;a = 2.5m/{s^2}]$
$⇒$ $u = \frac{{45}}{t} + 1.25\;t$
To find the minimum value of $u$
$\frac{{du}}{{dt}} = 0$ so we get $t = 6\sec$ then,
$u = \frac{{45}}{6} + 1.25 \times 6 = 7.5 + 7.5 = 15\,m/s$
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