Question

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

Step1: Recall acceleration - position relation

Acceleration $a(t)$ is the second - derivative of position $x(t)$ with respect to time $t$. Given $x(t)=3t^{2}-t$.

Step2: Find the first - derivative (velocity)

The derivative of $x(t)$ using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$ is $v(t)=\frac{dx}{dt}=\frac{d}{dt}(3t^{2}-t)=6t - 1$.

Step3: Find the second - derivative (acceleration)

Differentiate $v(t)$ with respect to $t$ to get $a(t)$. Using the power rule again, $a(t)=\frac{dv}{dt}=\frac{d}{dt}(6t - 1)=6$.

Answer:

$6$