Question

a particle moves along the x - axis so that at time t ≥ 0 its velocity is given by v(t)=-3t² + 12t + 36. determine the acceleration of the particle at t = 4.

Explanation:

Step1: Recall acceleration - velocity relationship

Acceleration $a(t)$ is the derivative of velocity $v(t)$. Given $v(t)=-3t^{2}+12t + 36$, we use the power - rule for differentiation $\frac{d}{dt}(t^{n})=nt^{n - 1}$.

Step2: Differentiate $v(t)$

$a(t)=v^\prime(t)=\frac{d}{dt}(-3t^{2}+12t + 36)=-3\times2t+12=-6t + 12$.

Step3: Evaluate $a(t)$ at $t = 4$

Substitute $t = 4$ into $a(t)$. $a(4)=-6\times4+12$.
$a(4)=-24 + 12=-12$.

Answer:

$-12$