Question

A man loses $20\%$ of his velocity after running through $108\text{m}$. If the rate by which he is losing his velocity remains constant, what is the maximum distance he will cover?

• 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 $u$. We know the man loses $20\%$ of his velocity after running $108\text{m}$.
Hence, velocity at this moment$=80\%\text{of u}=0.8u$
$\therefore$Final velocity $v=0.8u$
Also $s=108\text{m}$
Using $v^2=u^2+2as$,
We get,
$\Rightarrow (0.8u{)}^2=u^2+2a\times (108)$
$\Rightarrow 216a=0.64u^2-u^2=-0.36u^2$
$\Rightarrow a=-\frac{0.36}{216}{u}^2=-\frac{1}{600}{u}^2$

Now let's assume that the man ran a total distance of $x\text{m}$.
We know,
initial velocity $=u$
final velocity $v=0$
Distance covered $s=x$
Acceleration $a=-\frac{1}{600}{u}^2$
Using $v^2=u^2+2as$
We get,
$0=(u{)}^2+2ax$
$\to 0=u^2+2(-\frac{1}{600}{u}^2)x$
$\to \left(\frac{1}{300}{u}^2\right)x=u^2$
$\Rightarrow x=300\text{m}$
Hence total distance that the man can cover $=300\text{m}$