The stopping distance d of a car after the brakes are applied varies directly as the square of the speed r. if a car traveling 30 mph can stop in 50 ft, how many feet will it take the same car to stop when it is traveling 70 mph? (round to the nearest integer as needed)

d ≅ 272ft. A car that is traveling at 70mph it will stop 272ft after the brakes are applied.

The key to solve this problem is using the equation d = k(r²), where d is the distance after the brakes are applied, k is the desaceleration constant and r is the speed of the car.

In order to maintain the consistency of the units, we have to convert mph to ft/s using the equation ft/s = mph x 1.467.

30mph x 1.467 ≅ 44ft/s

We know the speed of the car and the distance travelled after brakes are applied. The, clear k for the equation d = k(r²)

k = d/(r²)

Solving with d = 50ft and r = 44 ft/s

k = 50ft/(44ft/s)²= 0.0258 s²/ft

Then, is the same car now is traveling at 70mph, how many feet will it take to stop?

Convert 70mph to ft/s

70mph x 1.467 = 102.69ft/s

Using the equation d = k(r²), where k = 0.0258 s²/ft and r = 102.69ft/s

d = 0.0258s²/ft[(102.69ft/s)²] = 272.06ft

109 is the nearest whole number/final answer! : )

maya’s expression will generate the wrong sum because she violated the commutative property.

step-by-step explanation: