Question

A man loses 20% of his velocity after running through $108\text{m}$. If his retardation is uniform, then the maximum total distance he can cover is

• A. $218\text{m}$
• B. $300\text{m}$
• C. $324\text{m}$
• D. $192\text{m}$

Answer 1

The correct option is B $300\text{m}$

Let the initial velocity be $u$. Since the man loses 20% of his velocity after running through $108\text{m}$,
Hence velocity at this moment = 80% of $u=0.8u$
Final velocity = $0.8u$ and $s=108\text{m}$
Using, $v^2=u^2+2as$
$(0.8u{)}^2=u^2+2a.(108)$
$216a=0.64u^2-u^2=-0.36u^2\Rightarrow a=-\frac{0.36}{216}{u}^2$
$a=-\frac{1}{600}{u}^2$
Now let the man run a further distance of $x\text{m}$. In second segment of his motion
Initial velocity $=0.8u$, final velocity $=0$, $s=x$
Using third equation of motion, $0=(0.8u{)}^2+2ax$ $0=0.64u^2+2(-\frac{1}{600}{u}^2)x$
$\Rightarrow \left(\frac{1}{300}{u}^2\right)x=0.64u^2$
$x=0.64\times 300=192\text{m}$
Hence total distance man can cover = $108+192=300\text{m}$