The Ideal Speed on a Banked Curve

Problem:

What is the ideal speed to take a 100 m radius curve banked at a 20.0° angle?

Solution:

The formula for the ideal speed on a banked curve can be derived from the formula of the ideal angle. That is, starting from \tan \theta = \frac{v^2}{rg}, we can solve for v.

v = \sqrt{rg \tan \theta}

For this problem, we are given the following values:

• radius of curvature, r=100\ \text{m}
• acceleration due to gravity, g=9.81\ \text{m/s}^2
• banking angle, \theta = 20.0 ^\circ

If we substitute the given values into our formula, we have

\begin{align*}
v = & \sqrt{rg \tan \theta} \\ \\
v = & \sqrt{\left( 100\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right) \left( \tan 20.0 ^\circ \right) } \\ \\
v = & 18.8959\ \text{m/s} \\ \\
v = & 18.9\ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The ideal speed for the given banked curve is about 18.9\ \text{m/s}.