##### 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
m/s

(b) the distance traveled
m

( c) the elapsed time
s
Jun 8th, 2015

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:

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.

Jun 8th, 2015

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Jun 8th, 2015
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Jun 8th, 2015
Dec 10th, 2016
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