A car starts from rest and moves with uniform acceleration ‘a’ on a straight road from time t = 0 to t=T. After that, a constant deceleration brings it to rest. In this process the average speed of the car is

Let the time the car is bought to rest is T₁.

After travelling for T seconds with an acceleration of a, the speed of the car is 'aT'.

So the average velocity of the car during deceleration will be given by  v'$=\frac{\mathrm{aT}+0}{2}$.

So the distance travelled will be $\frac{\mathrm{aT}}{2}.{\mathrm{T}}_1$

${\mathrm{v}}_{\mathrm{avg}}=\frac{\mathrm{Total} \mathrm{distance}}{\mathrm{total} \mathrm{time}}$

$\begin{array}{l}{\mathrm{v}}_{\mathrm{avg}}=\frac{\frac{1}{2}{\mathrm{aT}}^2+\frac{1}{2}\mathrm{aT}\times {\mathrm{T}}_1}{\mathrm{T}+{\mathrm{T}}_1}\\ =\frac{\frac{1}{2}\mathrm{aT}\left(\mathrm{T}+{\mathrm{T}}_1\right)}{\mathrm{T}+{\mathrm{T}}_1}\\ =\frac{1}{2}\mathrm{aT}\end{array}$