Question

question the position of a particle along a coordinate axis at time t (in seconds) is given, in meters, by s(t)=-2t^2 - 6t + 2. find the function that describes its acceleration at time t. provide your answer below: a(t)=□

Explanation:

Step1: Recall the relationship between position and acceleration

Acceleration $a(t)$ is the second - derivative of position $s(t)$. First, find the first - derivative of $s(t)=-2t^{2}-6t + 2$.
The derivative of $s(t)$ with respect to $t$ using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$ is $v(t)=s^\prime(t)=\frac{d}{dt}(-2t^{2}-6t + 2)=-4t-6$.

Step2: Find the second - derivative

The acceleration function $a(t)$ is the derivative of the velocity function $v(t)$.
Differentiate $v(t)=-4t - 6$ with respect to $t$. Using the power rule, $a(t)=v^\prime(t)=\frac{d}{dt}(-4t-6)=-4$.

Answer:

$a(t)=-4$