Question

What is the ideal speed to take a 105 m radius curve banked at a 30.0° angle? m/s

Answer 1

We know that the banked curve formula is given by:

`\begin{gathered} \tan\theta=(v^2)/(rg) \\ \text{ where} \\ \theta\text{ is the angle} \\ v\text{ is the velocity } \\ r\text{ is the radius } \\ g\text{ is the gravitational acceleration} \end{gathered}`

Plugging the values given and solving for v:

`\begin{gathered} \tan30=(v^2)/((105)(9.81)) \\ v^2=(105)(9.81)\tan30 \\ v=√((105)(9.81)\tan30) \\ v=24.4 \end{gathered}`

Therefore, the ideal speed is 24.4 m/s