Question

The velocity function of a car is given by v(t) = -3t2 + 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

Answer: In photo below

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Answer 1

The acceleration of the car three seconds before it comes to a stop is  -2.76 m/s² (Letter B).

Derivative

Derivative indicates the rate of change of a function with respect to a variable. Thus, when you derivate the position function, you find the velocity function. And, when you derivate the velocity function, you find the acceleration function.

In other words, the velocity function represents the first derivative of the position, meanwhile, the acceleration function represents the second derivative of the position.

For derivating an equation, you should apply the rule: $(\frac{d}{dx} ) (x^n ) = nx^{n-1}$. Example: x²= 2x

The car stops when the velocity is equal to zero. Thus, v(t) = -3t² + 18t + 9 =0. Therefore, you  should solve this quadratic function:

Δ=b²-4ac=$18^2-4\left(-3\right)\cdot \:9=324+12*9=324+108=432$

$t_{1,\:2}=\frac{-18\pm \sqrt{432}}{2\left(-3\right)}\\ \\ t_{1,\:2}=\frac{-18\pm \sqrt{432}}{-6}$

$t_1=\frac{-18+12\sqrt{3}}{-6}=3-2\sqrt{3}$

$t_2=\frac{-18-12\sqrt{3}}{-6}=3+2\sqrt{3}$

Like t is a positive number, you should use $t_2$. Thus, the value of t that you should apply to find the acceleration of the car three seconds before it comes to a stop is $2\sqrt{3}$.

The acceleration can be found from the derivative of the velocity function. See below.

a(t)= -6t+18 , for t=$2\sqrt{3}$

a(t)= -6*$2\sqrt{3}$t+18

a(t)= $-12\sqrt{3}$t+18

a(t)=-2.76

Read more about the derivative here:

brainly.com/question/29257155

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