Centripetal Acceleration?

Question

A race car starts from rest on a circular track of radius 492 m. The car's speed increases at the constant rate of 0.690 m/s2. At the point where the magnitudes of the centripetal and tangential accelerations are equal, find the following.
(a) the speed of the race car
(b) the distance traveled
(c) the elapsed time

Answer 1

v = wR; where v = tangential velocity, w = angular velocity (rad/sec), and R is the radius = 492 m.
dv/dt = a the tangential acceleration = .69 m/sec^2.

Centripetal/centrifugal force C = mv^2/R; where v = wR the tangential velocity given earlier.

(a) By inspection we see that alpha = v^2/R, which is the centripetal acceleration. So when a = .69 = v^2/R = alpha; we have v^2 = .69*492 = aR. You can do the math.

(b) We find distance traveled from the SUVAT equation v^2 = 2 aS; so that S = v^2/2a = aR/2a = R/2. You can do the math.

(c) Time elapsed can be found from the SUVAT v = at = sqrt(aR); so that t = sqrt(aR)/a = sqrt(R/a). You can do the math.

There really isn't much physics to be learned here. It's mostly recognizing that tangential acceleration a = centripetal acceleration = v^2/R; where v is the tangential velocity and R is the radius of turn. So under these very special conditions, when the two accelerations are equal, we can find some interesting relationships, like v^2 = aR. But given this is a very special case, it's not really that important. The rest of the problem is just invoking the right SUVAT equation.