The speed of a car traveling in a straight line is reduced from 55 to 40 mph in a distance of 362 feet. Find the distance in which the car can be brought to rest from 40 mph, assuming the same constant deceleration. (You must show the calculus behind your response)
a(t) = v'(t) = d''(t)
The Attempt at a Solution
Since acceleration is constant,
a(t) = c1
v(t) = c1*x + c2
d(t) = .5*c1*x^2 + c2*x + c3
I have no idea where to go from here.