Question

Civil engineers calculate suggested speeds around corners (those yellow signs) when designing roads based upon the ability of a car to remain upon the road. The ability of the car to remain upon the road when taking a corner is determined by the centripetal acceleration of your car around the corner. If you are driving a car around a flat circular turn with a suggested speed of 15 mph, how much more likely is it that your car does not successfully take the corner (ends up in the ditch) if you were to instead drive 45 mph? Cite an equation to back up your thinking.​

Answer 1

The car shall not be successful since the corner must have a radius as nine times as the real corner designed for a suggested speed of 15 miles per hour.

By definition of centripetal acceleration, the square of the velocity taken by the vehicle ($v$), in miles per hour, is directly proportional to the radius of the corner ($R$), in meters. Then, we have the following relationship:

$\frac{v_{A}^{2}}{R_{A}} = \frac{v_{B}^{2}}{R_{B}}$ (1)

Where:

• $v_{A}$ - Suggested velocity.
• $v_{B}$ - Real velocity.
• $R_{A}$ - Real radius of the corner.
• $R_{B}$ - Expected radius of the corner.

If we know that $v_{A} = 15\,\frac{mi}{h}$, $v_{B} = 45\,\frac{mi}{h}$ and $R_{A} = k$, then the expected radius of the corner is:

$R_{B} = k\cdot \left(\frac{v_{B}}{v_{A}} \right)^{2}$

$R_{B} = 9\cdot k$

In order to successfully take a corner at 45 miles per hour, the corner must have a radius as nine times as the real corner designed for a suggested speed of 15 miles per hour. Thus, the car shall not be successful at a speed of 45 miles per hour.

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