Question

question a particle travels along a horizontal line according to the function s(t)=t^4 - 8t^3 - 9t^2 + 2 where t is measured in seconds and s is measured in feet. find the acceleration function a(t). provide your answer below. a(t)=□

Explanation:

Step1: Recall the relationship between position, velocity and acceleration

Acceleration $a(t)$ is the second - derivative of the position function $s(t)$. First, find the first - derivative of $s(t)$ to get the velocity function $v(t)$. The power rule for differentiation is $\frac{d}{dt}(t^n)=nt^{n - 1}$.
$v(t)=\frac{d}{dt}(s(t))=\frac{d}{dt}(t^4-8t^3 - 9t^2 + 2)$
$v(t)=4t^{3}-24t^{2}-18t$

Step2: Differentiate the velocity function to get the acceleration function

Differentiate $v(t)$ with respect to $t$ using the power rule again.
$a(t)=\frac{d}{dt}(v(t))=\frac{d}{dt}(4t^{3}-24t^{2}-18t)$
$a(t)=12t^{2}-48t - 18$

Answer:

$a(t)=12t^{2}-48t - 18$