The stopping distance d of a car after the brakes have been applied varies directly as the square of the speed r. if the car traveling 70mph cab stop in 270ft, how many feet will it take the same car to stop when it 1s traveling 50 mph

Stopping distance = C r^2

270 = C*70^2
C = 270/70^2

At 50 mph,
stopping distance = 50^2*270/(70^2)

= (5/7)^2 * 270 ft = ?

To find out how many feet it will take for the car to stop when it is traveling at 50 mph, we can use the concept of direct variation between the stopping distance and the square of the speed.

We are given that the stopping distance (d) varies directly as the square of the speed (r). Mathematically, this can be represented as d = k * r^2, where k is the constant of variation.

First, we need to find the value of k. We can do this by using the information given that the car traveling 70 mph can stop in 270 ft. Plugging in these values into the formula, we get:

270 = k * (70)^2

To solve for k, divide both sides of the equation by (70)^2:

k = 270 / (70)^2

Next, we can use this value of k to find the stopping distance when the car is traveling at 50 mph. Plugging in these values into the formula, we get:

d = k * (50)^2

Now, substitute the value of k:

d = (270 / (70)^2) * (50)^2

Simplify and calculate the result:

d = (270 / 4900) * 2500
d = (27/49) * 2500
d ≈ 1367.35 ft

Therefore, it will take approximately 1367.35 feet for the car to stop when it is traveling at 50 mph.