A car, starting from rest, acceleration at the rate f through a distance S, the continues at constant speed for time t and then decelerates at the rate f/2 to come to rest. If the total distance travelled is 15S, then $\mathrm{S}=\frac{\mathrm{n}}{144}{\mathrm{ft}}^2$ then n value

The graph between velocity and time gives the distance travelled by the body in the motion. The velocity –time graph for the given situation can be drawn as below.

Magnitudes of slope of OA = f and slope of $\mathrm{BC}=\frac{\mathrm{f}}{2}$

$\begin{array}{l}\mathrm{V}={\mathrm{ft}}_1=\frac{\mathrm{f}}{2}{\mathrm{t}}_2\\ \therefore {\mathrm{t}}_2=2{\mathrm{t}}_1\end{array}$

In graph, area of $△$OAD gives distances, $\mathrm{S}=\frac{1}{2}{\mathrm{ft}}_1^2={\mathrm{S}}_1$     (say)         ........(i)

Area of rectangle ABCD gives distance travelled in time.

${\mathrm{S}}_2=\left({\mathrm{ft}}_1\right)\mathrm{t}$

Distance travelled in time

${\mathrm{t}}_3={\mathrm{S}}_3=\frac{1}{2}\frac{\mathrm{f}}{2}{\left(2{\mathrm{t}}_1\right)}^2={\mathrm{ft}}_1^2$

Thus, ${\mathrm{S}}_1+{\mathrm{S}}_2+{\mathrm{S}}_3=15\mathrm{S}$

$\mathrm{S}+\left({\mathrm{ft}}_1\right)\mathrm{t}+{\mathrm{ft}}_1^2=15\mathrm{S}$

$\mathrm{S}+\left({\mathrm{ft}}_1\right)+2\mathrm{S}=15\mathrm{S}\left(\mathrm{S}=\frac{1}{2}{\mathrm{ft}}_1^2\right)$

$\left({\mathrm{ft}}_1\right)\mathrm{t}=12\mathrm{S}$          ………(ii)

From Eqs. (i) and (ii), we have

$\begin{array}{l}\frac{12\mathrm{S}}{\mathrm{S}}=\frac{\left({\mathrm{ft}}_1\right)\mathrm{t}}{\frac{1}{2}\left({\mathrm{ft}}_1\right){\mathrm{t}}_1}\\ \therefore {\mathrm{t}}_1=\frac{\mathrm{t}}{6}\end{array}$

From Eq. (i), we gets $\mathrm{S}=\frac{1}{2}\mathrm{f}{\left({\mathrm{t}}_1\right)}^2$

$\therefore \mathrm{S}=\frac{1}{2}\mathrm{f}{\left(\frac{\mathrm{t}}{6}\right)}^2=\frac{1}{72}{\mathrm{ft}}^2$