You are told that
d(v) = kv^2
140 = k*2500
k = 0.056
so,
d(30) = .056*900
d(70) = .056*4900
The distance required for an emergency stop for a car varies directly as the square of the car’s speed. A car traveling at 50 mph requires 140 feet to stop. What is the stopping distance for a car traveling at 30 mph? What is the stopping distance for a car traveling at 70 mph?
3 answers
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