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

User Generated

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

Student has agreed that all tutoring, explanations, and answers provided by the tutor will be used to help in the learning process and in accordance with Studypool's honor code & terms of service.

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

Wbeqna O. (82)
UT Austin
Review
Review

Anonymous
Great! Studypool always delivers quality work.

Studypool
4.7
Trustpilot
4.5
Sitejabber
4.4