The 3210-lb car is traveling at 27 mi/hr when the driver applies the brakes, and the car continues to move along the circular path. What is the maximum deceleration possible (a; is negative if the speed is decreasing) if the tires are limited to a total horizontal friction force of 2110 lb? 110 ft Answer: a = i ft/sec?

Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter5: Newton's Laws Of Motion
Section: Chapter Questions
Problem 63PQ: A 75.0-g arrow, fired at a speed of 110 m/s to the left, impacts a tree, which it penetrates to a...
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The 3210-lb car is traveling at 27 mi/hr when the driver applies the brakes, and the car continues to move along the circular path.
What is the maximum deceleration possible (a; is negative if the speed is decreasing) if the tires are limited to a total horizontal
friction force of 2110 lb?
110 ft
Answer: a =
i
ft/sec2
Transcribed Image Text:The 3210-lb car is traveling at 27 mi/hr when the driver applies the brakes, and the car continues to move along the circular path. What is the maximum deceleration possible (a; is negative if the speed is decreasing) if the tires are limited to a total horizontal friction force of 2110 lb? 110 ft Answer: a = i ft/sec2
A Formula-1 car encounters a hump which has a circular shape with smooth transitions at either end. (a) What speed vg will cause
the car to lose contact with the road at the topmost point B? (b) For a speed va = 187 km/h, what is the normal force exerted by the
road on the 710-kg car as it passes point A?
B
11.1°-
p = 314 m
Answers:
(a) VB =
i
m/s
(b) NA =
i
N
Transcribed Image Text:A Formula-1 car encounters a hump which has a circular shape with smooth transitions at either end. (a) What speed vg will cause the car to lose contact with the road at the topmost point B? (b) For a speed va = 187 km/h, what is the normal force exerted by the road on the 710-kg car as it passes point A? B 11.1°- p = 314 m Answers: (a) VB = i m/s (b) NA = i N
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