If a car is traveling 50 mph downhill then its braking dista

If a car is traveling 50 mph downhill. then its braking distance on wet pavement is given by Evaluate D(-0.l) and interpret the result. What happens to braking distance as the downhill grade becomes steeper? The graph of D has a vertical asymptote at x = -0.3. Give the physical significance of this asymptote. Estimate the grade associated with a braking distance of 350 feet. Suppose r varies directly with the square of m and inversely with s. If r = 12 when m = 6 and 5 = 4, find r when m = 4 and s =10.

Solution

Part (a)

D(x) = 2500/30(0.3 + x)

D(-0.1) = 2500/30(0.3 + (-.1))

D(-0.1) = 416.67, this is the distance required to stop

Part (b)

A steeper grade is more negative, like -0.2. You would expect it to take longer to stop.

Part (c)

at x= -0.3, the car will not stop on the downhill, it keeps on moving at 50mph when the brakes are applied

Part (d)

D(x) = 2500/30(0.3 + x)

The grade associated with a braking distance of 350 feet

350 = 2500/30(0.3 + x)

350 = 250/3(0.3 + x)

(350*3)/250 = 1/(0.3 + x)

4.2 = 1/(0.3 + x)

0.3 + x=1/4.2

x=(1/4.2)-0.3

x=-0.0619