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 $f/2$ to come at 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. $S=ft$
• D. $S=\frac{1}{72}f{t}^2$

Answer 1

The correct option is D $S=\frac{1}{72}f{t}^2$

Distance traversed is $S$ then velocity attained will be $v=\sqrt{2fS}$ ....... (1)

so the next distance covered with this constant velocity will be $x=vt=\sqrt{2fS}\times t$

now distance covered in the time when it comes to rest with $retardation$ $f$ will be $y=\frac{v^2}{2\left(f/2\right)}=\frac{v^2}{f}=\frac{2fS}{f}=2S$

(see equation-1)

net distance is $S+x+y=S+t\sqrt{2fS}+2S=3S+t\sqrt{2fS}$

given that the net distance is $15S$ so $15S=3S+t\sqrt{2fS}$

or $(12S{)}^2=t^2\times 2fS$ so $S=\frac{1}{72}f{t}^2$