Question

let (s(t)=4t^{3}-48t^{2}+180t) be the equation of motion for a particle. find a function for the velocity. (v(t)=) where does the velocity equal zero? hint: factor out the gcf. (t =) and (t =) find a function for the acceleration of the particle. (a(t)=) question help: video message instructor submit question

Explanation:

Step1: Recall velocity - displacement relation

The velocity function $v(t)$ is the derivative of the position - function $s(t)$. Given $s(t)=4t^{3}-48t^{2}+180t$, using the power rule $\frac{d}{dt}(t^{n}) = nt^{n - 1}$, we have $v(t)=\frac{d}{dt}(4t^{3}-48t^{2}+180t)$.
$v(t)=4\times3t^{2}-48\times2t + 180=12t^{2}-96t + 180$.

Step2: Find when velocity is zero

Set $v(t)=0$, so $12t^{2}-96t + 180 = 0$. Factor out the greatest - common factor (GCF) which is 12: $12(t^{2}-8t + 15)=0$. Then divide both sides by 12 to get $t^{2}-8t + 15 = 0$. Factor the quadratic equation: $(t - 3)(t - 5)=0$. Solving for $t$, we get $t = 3$ and $t = 5$.

Step3: Recall acceleration - velocity relation

The acceleration function $a(t)$ is the derivative of the velocity function $v(t)$. Since $v(t)=12t^{2}-96t + 180$, using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $a(t)=\frac{d}{dt}(12t^{2}-96t + 180)=12\times2t-96=24t-96$.

Answer:

$v(t)=12t^{2}-96t + 180$
$t = 3$
$t = 5$
$a(t)=24t-96$