A race car starts from rest on a circular track of radius 578 m. The car's speed

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A race car starts from rest on a circular track of radius 578 m. The car's speed increases at the constant rate of 0.740 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 
Jun 8th, 2015

Thank you for the opportunity to help you with your question!

The tangential acceleration is 0.740 m/s². 

If you know vector calculus, take derivatives of the vector ( r sin(ωt), r cos(ωt) ) -- a vector which goes around in a circle at a constant speed -- to find the following relationships: 

r = radius(578m)
v = ω r 
a = ω² r. 

That proves that a = v² / r. (Your textbook might have a non-vector-calculus proof of that, but vector calculus is the easy way to do it.) 

This means that for (a), you want to find the speed v where: 

v² / r = 0.740 m/s². 

You know r. Solve for v. =427m/s^2

The total distance traveled? Well, you have to accelerate from 0 to the value in the previous problem (roughly 20 m/s). That takes a certain amount of time. How much time? [We're answering (c) first.] 

Well, what does 0.740 m/s² mean? It means at t=0s, you're going at speed 0. Then at t=1s, you're going at 0.740 m/s. Then at t = 2s, you're going at 1.48 m/s. Then at t = 3s, you're going at 2.22 m/s. And so on. 

Play with that math a little to find that you'll take about 40 seconds to go at the about 20 m/s. (You will probably do this more accurately than my mental math. Just divide v / (0.740 m/s² ) to get the number of seconds it takes you to reach that top speed.) 

How far do you go? Well, you're speeding up constantly, from 0 m/s to 20 m/s. That means that your average speed is going to be roughly 16 m/s. 

If I average 16 m/s traveling for 40 s, then I go 640 meters, right? Right. So that answers (b). And now everything is solved. 

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 8th, 2015

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