Velocity vs. time graph can be drawn as

Let the particle moves with constant acceleration for first $t_1$ seconds. Clearly it retards for last $t_1$ seconds.

Maximum speed of the particles is 

$v=u+at=0+5t_1=5t_1$

Area of velocity vs time graph gives displacement 

$\therefore$  displacement = area of traperion

$\Rightarrow \left(\mathrm{average} \mathrm{velocitys}\right)\times \left(\mathrm{total} \mathrm{time}\right)=\frac{1}{2}\left[25+\left(25-2t_1\right)\right]\times 5t_1$

$\Rightarrow 20\times 25=\frac{1}{2}\left(50-2t_1\right)\times 5t_1$

$\Rightarrow 1000=250t_1-10{t_1}^2$

$\Rightarrow 10{t_1}^2+250t_1-1000=0$

$\Rightarrow {t_1}^2+25t_1-100=0$

$\Rightarrow \left(t_1-5\right)\left(t_1-20\right)=0$

$t_1=5 \mathrm{or} t_1=20$

As  $t_1=20$ is not possible so $t_1=5 \mathrm{s}.$

$\therefore$ The car moved with uniform velocity for a time of $25-2t_1$

$=25-2\times 5$

$=15 \mathrm{s}.$