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College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Question
Determine the tangential and centripetal components of the net force exerted on the car (by the ground) in Example 5–8 when its speed is 15 m/s. The car’s mass is 950 kg.

Transcribed Image Text:* (I) Determine the tangential and centripetal components of the net force exerted on the
car (by the ground) in Example 5-8 when its speed is 15 m/s. The car's mass is 950 kg.

Transcribed Image Text:Example 5-8 Two components of acceleration
A race car starts from rest in the pit area and accelerates at a uniform rate to a speed of 35 m/s
in 11 s, moving on a circular track of radius 500 m. Assuming constant tangential acceleration,
find (a) the tangential acceleration, and (b) the radial acceleration, at the instant when the speed
is v = 15 m/s.
a. During the 11-s time interval, we assume the tangential acceleration atan is constant. Its
magnitude is
atan =
aR =
(35 m/s-0 m/s)
11 s
b. When v = 15 m/s, the centripetal acceleration is
=
= 3.2 m/s².
(15 m/s)²
(500 m)
= 0.45 m/s².
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