Car with tangential acceleration equal to centripetal acceleration

[jevans]

I need some help with a problem our professor handed out. Thanks in advance for any help.

A car is undergoing circular motion with a radius of 200m. Its initial speed is 3.00 m/s. The car moves so that its tangential acceleration is equal to its centripetal acceleration at all times. Calculate (a) its speed and (b) its centripetal acceleration 40.0 s later. (c) At what instant will the car's centripetal acceleration be equal to 9.80 m/s^2? (d) What will be its speed at that time.

The A sub t, and A sub c keeps tripping me up.

 

[Parth Dave]

Well I'll sort of try to start you off. First of all, you know that initially the centripetal acceleration is v^2/r = 9/.2 ms^-2. Hence, the initial tangential acceleration will be 45 ms^-2. Now, you'll also notice that the acceleration is not constant so your kinematics equations won't work. Your can come up with a formula for tangential acceleration based upon velocity. Also acceleration is dv/dt. So you can set up a differential equation. See where you can go with that.

 

[jevans]

Thanks for the help. I am struggling because my math skills haven't developed too far yet.

I worked out for part a) a sub c = dV/dT, so = (final V - 3)/40. This gave velocity of 4.8 m/s after 40 seconds.

From that b) is just a=V^2/R = 0.115 m/s^2

Part C) is the tricky part. Keep coming up with a sub c is dv/dt = v^2/r, then try to integrate. But I am not getting an answer that fits into the other #'s.

Part D) give acceleration of 9.8 = v^2/r, just solve for the v. 44.3 m/s.

Any other hints would be appreciated.