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

Answer: (C)

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 $t$

$x=\sqrt{2fS}\cdot ...(i)$ [As $s=ut$]

So the velocity of car at point $C$ also will be $\sqrt{2fs}$

and finally car stops after covering distance $y$.

Distance $CD$ i.e., $y=\frac{(\sqrt{2fS{)}^2}}{2(f/2)}=\frac{2fS}{f}=2S...(ii)$

[As $v^2=u^2-2as$

$\Rightarrow s=u^2/2a]$

So, total distance $AD=AB+BC+CD=15S$ (given)

$\Rightarrow S+x+2S=15S$

$\Rightarrow x=12S$

Substituting the value of $x$ in equation $(i)$ we get

$x=\sqrt{2fS}\cdot t$

$\Rightarrow 12S=\sqrt{2fS}\cdot t$

$\Rightarrow 144S^2=2fS\cdot t^2$

$\Rightarrow S=\frac{1}{72}f{t}^2$.