Question

A car, starting from rest, accelerates at the rate f through a distance S, then continuous 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 15 s, 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 car starts from point A from rest and moves up to point B with acceleration f.


Velocity of car at point B,
$v=\sqrt{2fs}$
$[As{v}^2=u^2+2as]$

Car moves distance BC with this constant velocity in time t.
$x=\sqrt{2fs}.t....\left(i\right)[Ass=ut]$
So the velocity of car at point C also will be $\sqrt{2fs}$ and finally car stops after covering distance y.

Distance $CD\Rightarrow y=\frac{(\sqrt{2fs}{)}^2}{2\left(\frac{f}{2}\right)}=\frac{2fs}{f}=2S...\left(ii\right)[As{v}^2=u^2-2as\Rightarrow s=\frac{u^2}{2a}]$

So, the total distance AD = AB + BC + CD = 15 S (given)
$\Rightarrow S+x+2S=15S\Rightarrow x=12S$
Substituting the value of x in equation (i) we get

$x=\sqrt{2fs}.t\Rightarrow 12S=\sqrt{2fs}.t\Rightarrow 144S^2=2fs.t^2$
$\Rightarrow S=\frac{1}{72}f{t}^2.$