Questions: The stopping distance d of a car after the brakes have been applied varies directly as the square of the speed r. If a car traveling 40 mph can stop in 80 ft, how fast can a car travel and still stop in 320 ft?

The stopping distance d of a car after the brakes have been applied varies directly as the square of the speed r. If a car traveling 40 mph can stop in 80 ft, how fast can a car travel and still stop in 320 ft?

Transcript text: The stopping distance d of a car after the brakes have been applied varies directly as the square of the speed r. If a car traveling 40 mph can stop in 80 ft, how fast can a car travel and still stop in 320 ft?

Solution

Solution Steps

Step 1: Establish the Relationship

The stopping distance \( d \) of a car varies directly as the square of the speed \( r \). This relationship can be expressed as: \[ d = k \cdot r^2 \] where \( k \) is the proportionality constant.

Step 2: Determine the Proportionality Constant

Using the given values, where a car traveling at \( r_1 = 40 \) mph stops in \( d_1 = 80 \) ft, we can find \( k \): \[ k = \frac{d_1}{r_1^2} = \frac{80}{40^2} = \frac{80}{1600} = 0.05 \]

Step 3: Calculate the New Speed

We need to find the speed \( r_2 \) at which a car can stop in \( d_2 = 320 \) ft. Using the relationship: \[ d_2 = k \cdot r_2^2 \] Substitute the known values: \[ 320 = 0.05 \cdot r_2^2 \] Solving for \( r_2 \): \[ r_2^2 = \frac{320}{0.05} = 6400 \] \[ r_2 = \sqrt{6400} = 80 \]