Drive Fast!

Let us consider the $v-t$ graph as shown. Let for the first t seconds it accelerates & for the next t seconds it decelerates.

Velocity in time t = $at$ The required distance = Area of $\Delta$ POE = 500

$\frac{1}{2}(at)(t+t)=500$

Solving we get total time of journey = $2\sqrt{\frac{500}{5}} = 2.10=20$

Thus minimum time is $\boxed{20}$ $\boxed{\mathfrak{seconds}}$

The maximum acceleration of the car is $5m/s^2$ . So acceleration is rate of change in velocity $dv/dt = 5$ thus by integration $v=5t$ and velocity is rate of change in displacement: $dx/dt = 5t$ so by repeating the same method $x= 5/2t^2$ . But you must end in zero velocity at the destination so the least amount of time would be speeding up to halfway and slowdown at the end.

Let t be time to reach halfway
$250 = 5/2 t^2$
$100 = t^2$
$t=10$
$2t=20$
Therefore the total time is 20 seconds

Half path=250=5/2.t^2. So,t=10, so it takes 10 sec.s to reach the half way and thus 20 sec. Is the ans.