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 rate $\frac{f}{2}$ to come to rest. If the total distance covered is $15s$, then which one is correct?

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

Answer 1

The correct option is A $s=\frac{f{t}^2}{72}$

When the car starts from rest and accelerates at the rate $f$
Distance covered is given by,
$\begin{array}{}s=0+\frac{1}{2}f{t}_1^2& \text{(1)}\end{array}$
$\Rightarrow t_1=\sqrt{\frac{2s}{f}}$
Velocity of the car after time $t_1$ is
$v=0+f{t}_1$
$v=f{t}_1$
Distance covered with constant speed $v$ in time $t$ is given by
$\begin{array}{}{s}^{\prime }=vt=\left(f{t}_1\right)t& \text{(2)}\end{array}$
Velocity of the car after time $t_2$ is 0 and let the distance covered be $s^{\prime \prime }$
$v^2=u^2+2ax$
$\Rightarrow 0=v^2-2\left(\frac{f}{2}\right)s^{\prime \prime }\Rightarrow v^2=f{s}^{\prime \prime }$
$\begin{array}{}\Rightarrow s^{\prime \prime }=\left(\frac{v^2}{f}\right)=\frac{f^2{t}_1^2}{f}=f{t}_1^2& \text{(3)}\end{array}$
We know that $s+s^{\prime }+s^{\prime \prime }=15s$
$\Rightarrow s^{\prime }+s^{\prime \prime }=14s$
Using (1), (2) and (3) we get
$\Rightarrow \left(f{t}_1\right)t+f{t}_1^2=14\text{s}$
$\Rightarrow \left(f{t}_1\right)t+2s=14s\Rightarrow \left(ft\right)t_1=12s$
$\therefore s=\left(\frac{ft}{12}\right)t_1=\left(\frac{ft}{12}\right)\sqrt{\frac{2s}{f}}\Rightarrow s^2=\frac{f^2{t}^2}{144}\times \frac{2s}{f}\Rightarrow s=\left(\frac{f{t}^2}{72}\right)$