3. The stopping distance \( d \) (in ft ) for a car on Mulholland Drive is given by \( d(v)=0.012 v^{2}+1.8 v \), where \( v \) is the speed of the car in mph the instant before the brakes were applied. (a) If the car was traveling 50 mph before the brakes were applied, find the stopping distance. (b) If the stopping distance is 91.2 ft , how fast was the car traveling before the brakes were applied?

To solve the problem, we will use the given stopping distance formula: \[ d(v) = 0.012 v^2 + 1.8 v \] where \( d \) is the stopping distance in feet and \( v \) is the speed of the car in mph. ### Part (a) We need to find the stopping distance when the car was traveling at 50 mph. We will substitute \( v = 50 \) into the formula: \[ d(50) = 0.012(50^2) + 1.8(50) \] Calculating \( 50^2 \): \[ 50^2 = 2500 \] Now substituting this value into the equation: \[ d(50) = 0.012(2500) + 1.8(50) \] Calculating \( 0.012 \times 2500 \): \[ 0.012 \times 2500 = 30 \] Calculating \( 1.8 \times 50 \): \[ 1.8 \times 50 = 90 \] Now, adding these two results together: \[ d(50) = 30 + 90 = 120 \] Thus, the stopping distance when the car was traveling at 50 mph is: \[ \boxed{120 \text{ ft}} \] ### Part (b) Now we need to find the speed \( v \) when the stopping distance \( d \) is 91.2 ft. We set up the equation: \[ 0.012 v^2 + 1.8 v = 91.2 \] Rearranging this gives us a standard quadratic equation: \[ 0.012 v^2 + 1.8 v - 91.2 = 0 \] To simplify the equation, we can multiply through by 1000 to eliminate the decimal: \[ 12 v^2 + 1800 v - 91200 = 0 \] Now we can use the quadratic formula \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 12 \), \( b = 1800 \), and \( c = -91200 \). Calculating the discriminant \( b^2 - 4ac \): \[ b^2 = 1800^2 = 3240000 \] \[ 4ac = 4 \times 12 \times (-91200) = -4368000 \] \[ b^2 - 4ac = 3240000 + 4368000 = 7608000 \] Now we can find \( v \): \[ v = \frac{-1800 \pm \sqrt{7608000}}{2 \times 12} \] Calculating \( \sqrt{7608000} \): \[ \sqrt{7608000} \approx 2755.0 \] Now substituting back into the formula: \[ v = \frac{-1800 \pm 2755.0}{24} \] Calculating the two possible values for \( v \): 1. Using the positive root: \[ v = \frac{-1800 + 2755.0}{24} = \frac{955.0}{24} \approx 39.79 \] 2. Using the negative root (which we discard since speed cannot be negative): \[ v = \frac{-1800 - 2755.0}{24} \text{ (not valid)} \] Thus, the speed of the car before the brakes were applied, when the stopping distance is 91.2 ft, is approximately: \[ \boxed{39.79 \text{ mph}} \]