Circular Motion Help: Solve for A sub t, A sub c, Speed & More

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SUMMARY

The discussion focuses on solving a physics problem involving circular motion, specifically calculating the speed and centripetal acceleration of a car with a radius of 200m and an initial speed of 3.00 m/s. The car's tangential acceleration equals its centripetal acceleration throughout the motion. After 40 seconds, the speed is determined to be 4.8 m/s, and the centripetal acceleration is calculated as 0.115 m/s². Additionally, the time when the centripetal acceleration reaches 9.80 m/s² corresponds to a speed of 44.3 m/s.

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jevans
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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.

:smile:
 
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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.
 
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.
 

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