A car travels at a constant speed around a circular track whose radius is 2.5 km. The car goes once around the track in 430 s. What is the magnitude of the centripetal acceleration of the car?
The centripetal acceleration is
a = V^2/R,
and the velocity V is given by
2*pi*R = V*P
where P is the period of 430 s.
The second equation can be used to eliminate the variable V, leaving you with
a = [2 pi R/P]^2/R = 4 pi^2 R/P^2
To find the magnitude of the centripetal acceleration of the car, we can use the formula:
a = (v^2) / r
Where:
a is the centripetal acceleration
v is the velocity of the car
r is the radius of the circular track
To find the velocity, we can use the formula:
v = circumference / time
The circumference of a circle can be found using the formula:
circumference = 2πr
Substituting these formulas, we have:
v = (2πr) / t
Given that the radius of the circular track is 2.5 km and the car goes once around the track in 430 s, we can plug in these values:
v = (2π * 2.5) / 430
v ≈ 0.36 km/s
Now, we can substitute the values for v and r in the formula for centripetal acceleration:
a = (0.36^2) / 2.5
a ≈ 0.0518 km/s^2
Therefore, the magnitude of the centripetal acceleration of the car is approximately 0.0518 km/s^2.