A car travels at a constant speed around a circular track whose radius is 2.87 km. The car goes once around the track in 332 s. What is the magnitude of the centripetal acceleration of the car?
2 pi radians in 332 sec
so omega = angular velocity = pi/166
Ac = omega^2 r
= (pi/166)^2 * 2870
To find the magnitude of the centripetal acceleration of the car, we can use the formula:
\[a = \frac{v^2}{r}\]
where:
- \(a\) is the centripetal acceleration,
- \(v\) is the velocity, and
- \(r\) is the radius of the circular track.
From the given information, we know that the radius of the circular track is 2.87 km. To find the velocity of the car, we can use the formula:
\[v = \frac{2\pi r}{t}\]
where:
- \(v\) is the velocity,
- \(r\) is the radius of the circular track, and
- \(t\) is the time taken to go once around the track.
Plugging in the values, we have:
\[v = \frac{2\pi \times 2.87 \, \text{km}}{332\, \text{s}}\]
To simplify the calculation, let's convert the radius from kilometers to meters:
\[v = \frac{2\pi \times 2870 \, \text{m}}{332\, \text{s}}\]
Now, we can calculate the velocity:
\[v = \frac{2 \times 3.1416 \times 2870}{332} \, \text{m/s}\]
Using a calculator, we find that \(v \approx 25.787 \, \text{m/s}\).
Now we can substitute the values of \(v\) and \(r\) into the centripetal acceleration formula:
\[a = \frac{(25.787 \, \text{m/s})^2}{2870 \, \text{m}}\]
Calculating this expression, we find:
\[a \approx 0.232 \, \text{m/s}^2\]
Therefore, the magnitude of the centripetal acceleration of the car is approximately \(0.232 \, \text{m/s}^2\).