An accepted relationship between stopping distance, d in feet, and the speed of

Question

An accepted relationship between stopping distance, d in feet, and the speed of a car, in mph, is d(v) = 1.2v + 0.06v^2 on dry, level concrete. (a) How many feet will it take a car traveling 45 mph to stop on dry, level concrete? (b) If an accident occurs 250 feet ahead, what is the maximum speed at which one can travel to avoid being involved in the accident? (a) It will take ft to stop the car that is traveling at 45 mph. (Round up to the nearest foot as needed.) (b) The maximum speed is mph at which one can travel to avoid being involved in the accident. (Round down to the nearest mile per hour as needed.)

Explanation / Answer

A.

d(v) = 1.2*v + 0.06*v^2

we need to find d, when v = 45 mph

d(v) = 1.2*45 + 0.06*45^2

d(v) = 175.5 ft

It will take 175.5 ft to stop the car that is travelling at 45 mph

2.

Now given that accedent will occur at 250 ft, so max speed should be

250 = 1.2*v + 0.06*v^2

0.06*v^2 + 1.2*v - 250 = 0

Solving above quadratic equation

v = [-1.2 +/- sqrt (1.2^2 + 4*0.06*250)]/(2*0.06)

v1 = 55.32 mph

v2 = -75.32 mph

since speed cannot be negtive, so the maximum speed is 55.32 mph

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