A car starts moving rectilinearly, first with acceleration $\alpha =5 m/s^2$ (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate $\alpha$ comes to a stop. The total time of motion equals $\tau =25 s$. The average velocity during that time is equal to $<v>=72 km/h$. How long does the car move uniformly?

Let $t$ be the time upto which car accelerates or decelerates. The maximum velocity attained in this duration is $5 t$. The time upto which car move uniformly $=25-2t$. The velocity - time graph of the motion of car is drawn as in figure.

Given the average velocity in whole time of motion

                              $v_{av}=\frac{72\times 5}{18}=20 m/s$.

The average velocity from the graph can be obtained as

                             $$\begin{gathered} v_{av}=\frac{total displacement}{total time} \\ =\frac{area of \overrightarrow{v}-t graph}{total time} \\ \therefore 20=\frac{{\displaystyle \frac{1}{2}}\times \left[25+\left(25-2t\right)\right]\times 5t}{25} \\ =\frac{{\displaystyle \frac{1}{2}}\times \left[50-2t\right]\times 5t}{25} \end{gathered}$$

                  or                $200=50t-2t^2$

                  or $t^2-25t+100=0$

                       $\left(t-20\right)\left(t-5\right)=0$

                                          $t=5 s$    or    $20 s$

But $t=20$ is not possible

                                    $t=5s$.

The time upto which car moves uniformly

                                      $$\begin{gathered} =25-2t=25-2\times 5 \\ =15 s. \end{gathered}$$