To find the speed that results in a braking distance of 120 feet, we start with the equation given:
[tex]\[ d = s + \frac{s^2}{10} \][/tex]
Here, \( d \) represents the braking distance, and \( s \) represents the speed. We need to find \( s \) when \( d = 120 \) feet. Thus, we set up the equation as follows:
[tex]\[ 120 = s + \frac{s^2}{10} \][/tex]
Next, we want to solve this equation for \( s \). To do this, we first multiply all terms by 10 to clear the fraction:
[tex]\[ 10 \times 120 = 10s + s^2 \][/tex]
[tex]\[ 1200 = 10s + s^2 \][/tex]
This rearranges to the standard quadratic form:
[tex]\[ s^2 + 10s - 1200 = 0 \][/tex]
We now need to solve the quadratic equation:
[tex]\[ s^2 + 10s - 1200 = 0 \][/tex]
To find the solutions, we can use the quadratic formula:
[tex]\[ s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where \( a = 1 \), \( b = 10 \), and \( c = -1200 \). Plugging these values into the formula, we get:
[tex]\[ s = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot (-1200)}}{2 \cdot 1} \][/tex]
[tex]\[ s = \frac{-10 \pm \sqrt{100 + 4800}}{2} \][/tex]
[tex]\[ s = \frac{-10 \pm \sqrt{4900}}{2} \][/tex]
[tex]\[ s = \frac{-10 \pm 70}{2} \][/tex]
This results in two potential solutions for \( s \):
[tex]\[ s = \frac{-10 + 70}{2} = \frac{60}{2} = 30 \][/tex]
or
[tex]\[ s = \frac{-10 - 70}{2} = \frac{-80}{2} = -40 \][/tex]
Since speed cannot be negative, we discard the solution \( s = -40 \). Therefore, the valid speed that results in a braking distance of 120 feet is:
[tex]\[ s = 30 \text{ miles per hour} \][/tex]
