Centripetal Acceleration and Tangential Acceleration problem driving me crazy

[roflawlz]

A car is tested on a 230 m diameter track.
If the car speeds up at a steady 1.2 m/s^2, how long after starting is the magnitude of its centripetal acceleration equal to the tangential acceleration?

it seems simple, but i just can't seem to get it!...
find: change in time when centripetal acceleration = tangential acceleration.
r = 115 m.
a = 1.2 m/s^2 (which type of a, i don't know)
centripetal accel = v^2/r or w^2/r
tangential accel = (radius)(angular accel)

i tried setting them equal to each other obviously, but no luck. can someone help me out??

 

[spork]

>> centripetal accel = v^2/r or w^2/r

centripetal accel = v^/r NOT w^2/r


>> tangential accel = (radius)(angular accel)

Nope, tangential accel is simply the acceleration given in the problem statement - 1.2 m/s^2

What speed must the car be going to give it a centripetal acceleration equal to 1.2 m/s^2 ? How long will it take to reach this speed?

Make sense?

 

[baba89]

Firstly, it is important to clarify the type of acceleration being referred to. In this problem, the centripetal acceleration is the acceleration towards the center of the circular track, while the tangential acceleration is the acceleration along the tangent of the track.

To solve this problem, we can use the equations for centripetal and tangential acceleration. The centripetal acceleration can be calculated using the formula a = v^2/r, where v is the velocity and r is the radius of the circle. The tangential acceleration can be calculated using the formula a = rw^2, where r is the radius and w is the angular velocity.

Since we are given the values for r and a (1.2 m/s^2), we can equate the two equations to find the velocity (v) at which the centripetal acceleration is equal to the tangential acceleration.

1.2 m/s^2 = v^2/r = (rw)^2/r = rw^2

Solving for w, we get w = √(1.2/r). Now, we can use the formula for angular velocity to find the time (t) it takes for the car to reach this velocity.

w = √(1.2/r) = ωf - ωi/t
where ωf is the final angular velocity and ωi is the initial angular velocity (which is 0).

Therefore, t = ωf / √(1.2/r) = (2πf) / √(1.2/r)
where f is the frequency of the car's motion.

We can also use the formula for tangential acceleration to find the time:

a = rw^2 = r(ωf-ωi)^2/t = r(2πf)^2/t
Solving for t, we get t = (r(2πf)^2)/a.

Therefore, the magnitude of the centripetal acceleration will be equal to the tangential acceleration after a time of t = (r(2πf)^2)/a = (r(2πf)^2)/(1.2 m/s^2).

In this case, since we are given the diameter of the track (230 m), the radius (r) is half of this value (115 m). So, the time it takes for the magnitude of the centripetal acceleration to equal the tangential acceleration is t =