What is the magnitude of the centripetal acceleration? just need someone to check my work?

Question

A car travels at a constant speed around a circular track whose radius is 2.6 km. The car goes once around the tract in 360s. What is the magnitude of the centripetal acceleration of the car?


Ok so
r=2.6 km
t=360 s

ac=v^2/r

so first we find the velocity with 2pir/T = 2pi(2.6)/360s=.045m/s

then to find the ac its .045^2/2.6=7.78x10-4 m/s^2?

Answer 1

I think you should change your r to 2600 instead of 2.6

Answer 2

You worked it OK, but your numbers are different from what I got.

You might note that w = 2 pi/t; where w is the angular velocity of the car around the track in time t = 360 sec. In which case, tangential velocity = v = wR; where R = 2.6 km, so that a = v^2/R = (wR)^2/R = w^2 R = (2 pi/t)^2 R = .79 m/sec^2

Answer 3

it's right if the unit of acceleration is in km/s^2

V = 2pi r / T
V = 2(3.14)(2.6km) / 360s
V = .0454 km/s

a = v^2 / r

a = (.0454km/s)^2 / 2.6km

a = (.00206116 km^2 / s^2) / 2.6km

a = 7.9275 x 10^-4 km/s^2

Acceleration, in physics, is usually in m/s^2.

7.9275 x 10^-4 km/s^2 * (1000m / km) = .79275 m/s^2