The velocity function of a car is given by v(t) = -3t^2 + 18t + 9 m/s. Find the acceleration of the car three seconds before it comes to a stop. A. a = -6.46 m/s² B. a = -2.76 m/s² C. a = -0.46 m/s² D. a = 2.76 m/s²

The acceleration when the car comes to a full stop is 20.76 meters per square second.How to find the acceleration?Here we have the velocity function:v(t) = -3t^2 + 18t + 9To find the acceleration we need to differentiate with respect to the variable t, so we will get the acceleration:a(t) = 2*(-3)*t + 1*18a(t) = - 6*t + 18Now, the car will come to a stop when the velocity is equal to zero, so we need to solve:-3t^2 + 18t + 9 = 0Dividing all by -3 we get:(-3t^2 + 18t + 9)/-3 = 0t^2 - 6t - 3 = 0The solutions are:[tex]t = \frac{6 \pm \sqrt{(-6)^2 - 4*1*-3} }{2*1} \\\\t = \frac{6 \pm 6.92}{2}[/tex]We only care for the positive solution:t = (6 + 6.92)/2 = 6.46The acceleration at that time is:a(6.46) =  - 6*6.46 + 18 = -20.76Learn more about acceleration:https://brainly.com/question/14344386#SPJ1