Question

The function d(v) = 0.0067 v^2 + 0.15v can be used to determine the safe stopping distance, d(v), in metres for a car given its speed, v, in kilometres per house. determine the speed at which a car can be traveling in order to be able to stop at a distance of 24m. show your work and round your final answer(s) to the nearest meter.

Answer 1

Evaluate the function when d(v) = 24.

$\begin{gathered} 24=0.0067v^2+0.15v \\ 0.0067v^2+0.15v-24=0 \end{gathered}$

Use the quadratic formula to find v.

$v=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

Where a = 0.0067, b = 0.15, and c = -24.

$\begin{gathered} v=\frac{-0.15\pm\sqrt[]{0.15^2-4\cdot0.0067\cdot(-24)}_{}}{2\cdot0.0067}=\frac{-0.15\pm\sqrt[]{0.6657}_{}}{0.0134} \\ v_1=49.7\cdot\frac{km}{hr}_{} \\ v_2=-72.1\cdot\frac{km}{hr} \end{gathered}$

Therefore, the car should travel at 49.7 km/hr.

We take the positive speed as the answer.