A car travels at a constant speed around a circular track whose radius is 3.94 km. The car goes once around the track in 260s. What is the magnitude of the centripetal acceleration of the car?
The formula for centripetal acceleration is given by:
\[a = \frac{{v^2}}{r}\]
where \(v\) is the velocity and \(r\) is the radius.
To find \(v\), we can use the formula for linear velocity:
\[v = \frac{{2\pi r}}{T}\]
where \(T\) is the time period.
Plugging in the values:
\[v = \frac{{2\pi(3.94\, \text{km})}}{260\, \text{s}} \approx 0.4807\, \text{km/s}\]
Now we can plug in the values for \(v\) and \(r\) to calculate the centripetal acceleration:
\[a = \frac{{(0.4807\, \text{km/s})^2}}{3.94\, \text{km}} \approx 0.0592\, \text{km/s}^2\]
Therefore, the magnitude of the centripetal acceleration of the car is approximately \(0.0592\, \text{km/s}^2\).