Problema Solution
for a car traveling on dry level pavement, reaction distance r and braking distance b (both in feet) are functions of speed x in miles per hour. r(x)=4x and b(x)-1/10x2
write a formula for stopping distance s(x)
evaluate s(50) and interpret the result.
What speed gives a stopping distance of 50 feet?
Answer provided by our tutors
Stopping distance = Reaction distance + Breaking distance
s(x) = r(x) + b(x)
r(x) = 4x
b(x) = (1/10)x^2
s(x) = 4x + (1/10)x^2
s(50) = 4*50 + (1/10)*50^2
s(50) = 450 ft
The stopping distance for a car traveling with 50 mph is 450 feet.
What speed gives a stopping distance of 50 feet?
We need to find x, x>0 such that s(x) = 50 that is:
4x + (1/10)x^2 = 50
........
........
x = 10 mph
The speed of 10 mph gives a stopping distance of 50 feet.