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 $15 \,S$, then

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

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
$x = \sqrt{2fS} \cdot t$
$\Rightarrow 12S = \sqrt{2fS} \cdot t$
$\Rightarrow 144S^2 = 2fS \cdot t^2$
$\Rightarrow S = \frac{1}{72} ft^2$.