The stopping distance for a car traveling 25 mph is 69.2 feet, and for a car traveling 40 mph it is 112 feet. The stopping distance in feet can be described by the equation \( \mathrm{y}=\mathrm{ax} 2+\mathrm{bx} \), where \( x \) is the speed in mph. (a) Find the values of a and b . (b) Use the answers from part (a) to find the stopping distance for a car traveling 60 mph . \( \left[\begin{array}{ll|l}625 & 25 & 69.2 \\ 1600 & 40 & 112\end{array}\right] \) (Type whole numbers.) The value of a is \( \square \). The value of b is \( \square \). (TvDe inteaers or decimals rounded to six decimal places as needed.)

To find the values of \( a \) and \( b \) in the equation \( y = ax^2 + bx \), we can set up a system of equations based on the stopping distances given for 25 mph and 40 mph. We have: 1. For \( x = 25 \): \( 69.2 = a(25^2) + b(25) \) → \( 69.2 = 625a + 25b \) 2. For \( x = 40 \): \( 112 = a(40^2) + b(40) \) → \( 112 = 1600a + 40b \) This gives us the system of equations: 1. \( 625a + 25b = 69.2 \) (Equation 1) 2. \( 1600a + 40b = 112 \) (Equation 2) We can simplify the equations to make calculations easier. Dividing Equation 1 by 25, we get: \( 25a + b = 2.768 \) (Equation 1 simplified) Next, dividing Equation 2 by 40, we have: \( 40a + b = 2.8 \) (Equation 2 simplified) Now, we can solve this system: 1. From Equation 1 simplified, \( b = 2.768 - 25a \). 2. Substituting \( b \) into Equation 2 simplified: \( 40a + (2.768 - 25a) = 2.8 \) This simplifies to \( 15a + 2.768 = 2.8 \) or \( 15a = 0.032 \). Therefore, \( a = \frac{0.032}{15} = 0.0021333 \). Now substituting \( a \) back into Equation 1 simplified: \( 25(0.0021333) + b = 2.768 \), giving \( b = 2.768 - 0.053333 = 2.7146667 \) or approximately \( 2.714667 \). Now, we have: The value of \( a \) is approximately \( 0.002133 \) (if rounded to six decimal places). The value of \( b \) is approximately \( 2.714667 \) (if rounded to six decimal places). (b) To find the stopping distance for a car traveling 60 mph, we plug \( x = 60 \) into our equation: \( y = 0.002133(60^2) + 2.714667(60) \). Calculating: \( y = 0.002133(3600) + 2.714667(60) \), \( y = 7.6768 + 162.88002 = 170.55682 \). So the stopping distance for a car traveling 60 mph is approximately \( 170.56 \) feet.