A car, starting from rest, is accelerated at a constant rate $\mathrm{\alpha }$ until it attains a speed v. It is then retarded at a constant rate $\mathrm{\beta }$ until it comes to rest. The average speed of the car during its entire journey is:

The distance s₁ covered by the car when it is accelerated is given by $2{\mathrm{\alpha s}}_1={\mathrm{v}}^2$, which gives s₁ $=\frac{{\mathrm{v}}^2}{2\mathrm{\alpha }}$. 
The distance ${\mathrm{s}}_2$ covered when it is decelerated is, similarly given by ${\mathrm{s}}_2=\frac{{\mathrm{v}}^2}{2\mathrm{\beta }}$. 

Therefore, the total distance covered is:
$\mathrm{s}={\mathrm{s}}_1+{\mathrm{s}}_2=\frac{{\mathrm{v}}^2}{2}\left(\frac{1}{\mathrm{\alpha }}+\frac{1}{\mathrm{\beta }}\right)$                           …(i)
If t₁ is the time of acceleration and t₂ that of deceleration, then $\mathrm{v}={\mathrm{\alpha t}}_1={\mathrm{\beta t}}_2$ or ${\mathrm{t}}_1=\frac{\mathrm{v}}{\mathrm{\alpha }}$ and ${\mathrm{t}}_2=\frac{\mathrm{v}}{\mathrm{\beta }}$. Therefore, the total time taken is
$\mathrm{t}={\mathrm{t}}_1+{\mathrm{t}}_2=\mathrm{v}\left(\frac{1}{\mathrm{\alpha }}+\frac{1}{\mathrm{\beta }}\right)$                                 …(ii)

From (i) and (ii), the average speed of the car is given by
$\mathrm{average} \mathrm{speed} = \frac{\text{ total distance }}{\text{ total time }}=\frac{\mathrm{s}}{\mathrm{t}}=\frac{\mathrm{v}}{2}$