A car, starting from rest, has a constant acceleration a₁ for a time interval t₁ during which it covers a distance s₁. In the next time interval t₂, the car has a constant retardation a₂ and comes to rest after covering a distance s₂ in time t₂. Which of the following relations is correct? 

Since the initial velocity of the car is zero, its velocity at the end of the first time interval t₁ is v = 0 + a₁t₁= a₁t₁. This is the initial velocity for the next time interval t₂. Since the final velocity is zero, we have, from v = u + at,

0 = a₁t₁ - a₂t₂        ($\because$u=a₁t₁)

Now, the distance covered in the first time interval t₁ is given by 

$$\begin{gathered} 2a_1{s}_1=v^2-u^2=a_1^2{t}_1^2 \\ \left(\because v=a_1{t}_1\text{ and }u=0\right) \\ or s_1=\frac{1}{2}{a}_1{t}_1^2 \left(i\right) \end{gathered}$$

The distance covered in the next time interval t₂ is given by 

$$\begin{gathered} -2a_2{s}_2=0-a_1^2{t}_1^2 (\because v=0 and u=a_1{t}_1 now) \\ or s_2=\frac{1}{2}\frac{a_1^2}{a_2}{t}_1^2 \left(ii\right) \\ \mathrm{From} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get}: \\ \frac{{\displaystyle s_1}}{{\displaystyle s_2}}=\frac{{\displaystyle \frac{1}{2}{a}_1{t}_1^2}}{{\displaystyle \frac{1}{2}\frac{a_1^2}{a_2}{t}_1^2 }}=\frac{{\displaystyle a_2}}{{\displaystyle a_1}} \\ \because a_1{t}_1=a_2{t}_2 \\ \therefore \frac{a_1}{a_2}=\frac{t_2}{t_1} \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}: \frac{s_2}{s_1}=\frac{a_1}{a_2}=\frac{t_2}{t_1} \end{gathered}$$