Question

A car, starting from rest, accelerates at the rate $f$ through a distance $S$, then continues at constant speed for time $t$ and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed is $15S$, then

• A. $S=\frac{1}{2}f{t}^2$
• B. $S=\frac{1}{4}f{t}^2$
• C. $S=\frac{1}{72}f{t}^2$
• D. $S=\frac{1}{6}f{t}^2$

Answer 1

The correct option is C $S=\frac{1}{72}f{t}^2$

Let the value of car after covering $S$ distance by $x$. Then

By third equation of motion

$v^2-(0{)}^2=2fs\Rightarrow v=\sqrt{2fs}$

constant speed $v$ be $8$, then

$s^{\prime }=v.t\Rightarrow s=\left(\sqrt{2fs}\right)t$

While rerefraction, let the distance convered be $s^{\prime }$. By

By third equation of motion

$(0{)}^2-v^2=(2)\left(\frac{-f}{2}\right)s"$

$\Rightarrow s"=\frac{2fs}{f}$

$\Rightarrow s"=2s$

So, Total distance $\left(15s\right)=s+s^{\prime }+s"$

$\Rightarrow 15s=s+\sqrt{2fs}t+t2s$

$\Rightarrow (12{)}^2{s}^2=2f+t^{2}s\Rightarrow s=\frac{f{t}^2}{72}$

OPtion $C$ is correct.