A NASCAR racecar rounds one end of the Martinsville Speedway. This end of the tr

Question

A NASCAR racecar rounds one end of the Martinsville Speedway. This end of the track is a turn with radius approximately 187 ft. if we approximate the track to be completely flat and the racecar is traveling at a constant 60.00 mph around the turn, what is the racecar's centripetal (radial) acceleration? Convert this result to SI units. What is the force responsible for the centripetal acceleration in this case? To keep from skidding into the wall on the outside of the turn what is the minimum coefficient of static friction between the racecar's tires and track?

Explanation / Answer

given that radius of track   r = 187ft = 187 * 0.3048m = 56.99m    speed of car v = 66 mph    = 66 * 1.467 ft/s = 96.822 ft/s (or) v = 66 * 0.4470 m/s = 29.50 m/s centripetal acceleration   a = v2/r                                          = 50.13 ft/s2 ----------------------------
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the speed , v = rs g    v2/ rg = s = 1.5 the minimum coefficient of static friction between the racecar's tires and track is 1.5. the speed , v = rs g    v2/ rg = s = 1.5 = 1.5

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