​A race car starts from rest on a circular track of radius 284 m. The car's spee

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urfunzzntobby

Science

Question Description

A race car starts from rest on a circular track of radius 284 m. The car's speed increases at the constant rate of 0.410 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

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Explanation & Answer

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v^2/r=0.410

t=0s then at t=1 you're going 0.410m/s then at 2s its 0.82m/s and so on

v=root(.410*284)=10.7145

We can find the distance travelled by using constant acceleration equation: 
wf^2 = wi^2 + 2angular acceleration*(theta - theta initial) wf and wi are angular speeds final and initial respectively. theta and theta initial are angular displacement final and initial respectively. 

Since x = r*theta, v = r*w, a = r*(angular acceleration) 
w = v/r = 10.7145/284 =  .037727 rad/s 
ang acceler = 037727/284 = 1.32x10^-4 rad/s^2 

Using the equation above: 
037727^2 = 0^2 + 2*1.32x10^-4*theta 

0.001423326529

5.357 = theta 

x = r*theta, so the distance travelled is 
x =r* theta = 284m * 5.35 = 1521 m 
b) the car has travelled 1521 m when the speed is 10.7145 m/s 

c) time elapsed: 
theta = theta0 + wt + .5angular acceleration * t^2 
theta = .5at^2, since angular speed is zero initially and angular displacement is zero. 
root(2theta/a) = t 
t = root(2*5.35/1.32x10^-4) = 810.6 seconds


Please let me know if you need any clarification. I'm always happy to answer your questions.

Wbeqna O. (82)
UT Austin

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