An accepted relationship between stopping distance, d in feet, and the speed of a car, in mph, is d(v)=1.1v+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 200 feet ahead, what is the maximum speed at which one can travel to avoid being involved in the accident?

The given equation is:

d (v) = 1.1 v + 0.06 v^2

a. so given that v = 45 mph, find d in feet

d = 1.1 * 45 + 0.06 * 45^2

d = 171 feet

So it takes 171 feet before coming to a stop.

b. given d = 200 feet, calculate for v

200 = 1.1 v + 0.06 v^2

divide everything by 0.06:

v^2 + (1.1/0.06) v = (200/0.06)

completing the square:

(v + 1.1/0.03)^2 = (200/0.06) + (1.1/0.03)^2

v + 1.1/0.03 = ± 68.39

v = -105.06 mph; 31.72 mph

So the maximum speed is 31.72 mph

sorry can't

step-by-step explanation:

i googled though and something came up but it was in science langueage not common english sooo