Pedal to the Metal

Let the velocity of the race car be $v(t)$ , acceleration be $a(t)$ and jerk $j(t)$ . Then we have:

$\begin{aligned} j(t) & = \frac {da}{dt} = - 2 \\ \implies a(t) & = \int - 2 \ dt = -2t + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ a(t) & = -2t + 20 & \small \color{#3D99F6} \text{Since }a(0) = 20 \end{aligned}$

Then $v(t) = \displaystyle \int a(t)\ dt = 20t - t^2$ , since the car starts from rest. The car reaches its maximum is when $a(t) = 20-2t =0$ or $t=10$ and the maximum speed is $v(10) = 20(10)-10^2 = \boxed{100} \text{ m/s}$ .

Let $v''(t) = -2$ (i) be the racecar's jerk rate (with $v'(0) = 20, v(0) = 0$ as the initial acceleration and velocity respectively). Intergrating (i) once yields:

$v'(t) = -2t + A \Rightarrow v'(t) = -2t + 20$ (ii)

and integrating (ii) once gives:

$v(t) = -t^2 + 20t + B \Rightarrow v(t) = -t^2 + 20t = -(t-10)^2 + 100$ (iii)

The car's velocity is a concave-down parabola with its vertex at $(v,t) = (100, 10)$ . Hence, the maximum velocity is $\boxed{100}$ m/s.