Question

question: a particle travels along a horizontal line according to the function s(t)=t^3 - 3t^2 - 8t + 1 where t is measured in seconds and s is measured in feet. find the acceleration of the particle at t = 3 seconds. provide your answer below.

Explanation:

Step1: Recall the relationship between position, velocity and acceleration

The velocity $v(t)$ is the first - derivative of the position function $s(t)$, and the acceleration $a(t)$ is the second - derivative of the position function $s(t)$. Given $s(t)=t^{3}-3t^{2}-8t + 1$.

Step2: Find the first - derivative (velocity function)

Using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=s^\prime(t)=\frac{d}{dt}(t^{3}-3t^{2}-8t + 1)=3t^{2}-6t - 8$.

Step3: Find the second - derivative (acceleration function)

Differentiate $v(t)$ with respect to $t$. $a(t)=v^\prime(t)=s^{\prime\prime}(t)=\frac{d}{dt}(3t^{2}-6t - 8)=6t-6$.

Step4: Evaluate the acceleration at $t = 3$

Substitute $t = 3$ into the acceleration function $a(t)$. $a(3)=6\times3-6$.
$a(3)=18 - 6=12$ feet per second squared.

Answer:

$12$ feet per second squared