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 to rest. If the total distance traversed is 15 s, then s = ?

• 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 $s=\frac{f{t}^2}{72}$

If $t_1$ be the time for which the car accelerates at the rate $f$ from rest, the distance traveled in time $t_1$ is

$s=s_1=0\left(t_1\right)+\frac{1}{2}f{t}_1^2=\frac{1}{2}f{t}_1^2$ ....(1) (using formula

$s=ut+\frac{1}{2}a{t}^2$)

and the velocity at time $t_1$ is $v_1=f{t}_1$

So, the distance traveled in time $t$ will be $s_2=v_{1}t=f{t}_{1}t$ ...(2)

If $t_2$ be the time for which the car decelerates at the rate $f/2$ to come rest, the distance traveled in time $t_2$ is given by,

$0^2-v_1^2=2(-f/2)s_3$ (using formula $v^2-u^2=2as$)

or $s_3=(f{t}_1{)}^2/f=f{t}_1^2$ ...(3)

using (1), $s_3=2s_1=2s$

Given, $s_1+s_2+s_3=15s$

or $s+f{t}_{1}t+2s=15s$

or $ft{t}_1=12s$

or $12s=ft{t}_1$ ..(4)

$\left(4\right)/\left(1\right)\Rightarrow \frac{12s}{s}=\frac{ft{t}_1}{\left(1/2\right)f{t}_1^2}$ or $t_1=t/6$

From (1), $s=\left(1/2\right)f(t/6{)}^2=f{t}^2/72$