Assume the car travels to the right on an $x$-axis, and first applies the brakes at $x\=0$. Then, the motion of the car is described by $x\left(t\right)\=-30 \frac{t^2}{2}\+v_0 t$. (The calculus has been invoked by twice antidifferentiating $x″\left(t\right)\=-30$, and using the conditions $x\prime \left(0\right)\=v_0$ and $x\left(0\right)\=0$.)

The rest of the solution is algebraic. First, the total time taken to bring the car to a stop is found by solving $x\left(t\right)\=180$ for

$t\=\stackrel{\^}{t}\equiv \left(v_0\pm \sqrt{v_0^2-10800}\right)\/30$ 

The initial velocity can now be found by solving the equation $x\prime \left(\stackrel{\^}{t}\right)\=0$ for $v_0$. For either value of $\stackrel{\^}{t}$ this equation becomes

 $x\prime \left(\stackrel{\^}{t}\right)\equiv -30 \stackrel{\^}{t}\+v_0\=\pm \sqrt{v_0^2-10800}\=0$ 

from which it follows that $v_0\=\sqrt{10800}\=60\sqrt{3}\doteq 103.92$ ft/s.

Converting this to miles per hour gives $60\sqrt{3} \times \frac{3600}{5280}\doteq 70.86$ mph.

Converting this to kilometers per hour gives 114.033 km/h, where Maple's Units Calculator Assistant, as per Figure 3.10.5(a), was used.