The question:
When a car skids to a stop, the length L, in feet, of the skid marks is related to the speed S, in miles per hour, of the car by the power function below.
L = (1/30h)S^2
Here the constant h is the friction coefficient, which depends on the road surface.?For dry concrete pavement, the value of h is about 0.85. (Round your answers to two decimal places.)
(a) If a driver going 45 miles per hour on dry concrete jams on the brakes and skids to a stop, how long will the skid marks be?
79.41ft
(b) A policeman investigating an accident on dry concrete pavement finds skid marks 228 feet long. The speed limit in the area is 70 miles per hour. Is the driver in danger of getting a speeding ticket?
The driver's speed was 76.25 mph so it appears the driver would be in danger of getting a ticket.
(c) This part of the problem applies to any road surface, so the value of h is not known. Suppose you are driving at 55 miles per hour but, because of approaching darkness, you wish to slow to a speed that will cut your emergency stopping distance in half. What should your new speed be? (Hint: You should use the homogeneity property of power functions here. By what factor should you change your speed to ensure that L changes by a factor of 0.5?)
____ mphMath: when a car skids to a stop?you have already given the ans to qa %26amp; qb
they are straight applications of the forrmula
qc
let V mph be the speed that changes L by a factor of 0.5
L1 = (1/30h)55^2
L2 = 0.5L1 = 0.5*(1/30h)V^2
V^2/55^2 = 0.5
V = 55*sqrt(0.5)
= 38.9 mph
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