Question

the position of a particle along a coordinate axis at time t (in seconds) is given by s(t)=-3t^2 + 6t - 1 (in meters). find the function that describes its acceleration at time t. provide your answer below: a(t)=□ m/s^2

Explanation:

Step1: Recall the relationship between position, velocity and acceleration

Acceleration is the second - derivative of position function. First, find the velocity function $v(t)$ by differentiating the position function $s(t)=-3t^{2}+6t - 1$.
The power rule for differentiation is $\frac{d}{dt}(t^{n})=nt^{n - 1}$.
$v(t)=s^\prime(t)=\frac{d}{dt}(-3t^{2}+6t - 1)$
$v(t)=-3\times2t+6=-6t + 6$.

Step2: Differentiate the velocity function to get the acceleration function

Differentiate $v(t)=-6t + 6$ with respect to $t$ to find the acceleration function $a(t)$.
$a(t)=v^\prime(t)=\frac{d}{dt}(-6t + 6)$.
Using the power rule, $\frac{d}{dt}(-6t)=-6$ and $\frac{d}{dt}(6)=0$. So $a(t)=-6$.

Answer:

$a(t)=-6$