Strange Car

Let the acceleration of the strange car be $a(t)$ , then

$\begin{aligned} a(t) & = 30 - k(t-6)^2 & \small \color{#3D99F6} \text{where }k \text{ is a constant.} \\ a(0) & = 30 - k(-6)^2 = 6 & \small \color{#3D99F6} \text{Given that } a(0) = 6 \text{m/s}^2 \\ \implies k & = \frac {24}{36} = \frac 23 \\ \implies a(t) & = 30 - \frac 23 (t-6)^2 & \small \color{#3D99F6} \implies a(t) = 0 \text{ when }t = \sqrt{45} + 6 \text{ s} \\ & = 6 + 8t - \frac 23 t^2 \end{aligned}$

Now speed $v(t)$ and displacement $s(t)$ of the strange car is given as follows:

$\begin{aligned} v(t) & = \int a(t) \ dt = 6t + 4t^2 - \frac 29 t^3 & \small \color{#3D99F6} \text{Since }v(0) = 0 \\ s(t) & = \int v(t) \ dt = 3t^2 + \frac 43 t^3 - \frac 1{18} t^4 & \small \color{#3D99F6} \text{Since }s(0) = 0 \end{aligned}$

$v(t)$ is maximum when $a(t) = 0$ or $t = \sqrt {45} + 6$ seconds, then the distance traveled by the strange car is $s(\sqrt{45} +6) \approx \boxed{1.772}$ .