a particle moves along the x - axis so that at time t ≥ 0 its position is given by x(t)=-5t^4 + 30t^2…

Explanation:

Step1: Find the velocity function

The velocity function $v(t)$ is the derivative of the position - function $x(t)$. Using the power rule $\frac{d}{dt}(t^n)=nt^{n - 1}$, if $x(t)=-5t^{4}+30t^{2}$, then $v(t)=x^\prime(t)=-20t^{3}+60t$.

Step2: Set up the inequality for moving to the right

A particle is moving to the right when $v(t)>0$. So we need to solve the inequality $-20t^{3}+60t>0$. First, factor out $-20t$: $-20t(t^{2}-3)>0$, or $20t(t^{2}-3)<0$ (after multiplying both sides by - 1 and reversing the inequality sign). Further factor $t^{2}-3=(t-\sqrt{3})(t + \sqrt{3})$. So we have $20t(t-\sqrt{3})(t+\sqrt{3})<0$, and since $t\geq0$, we only consider $t(t - \sqrt{3})<0$.

Step3: Find the critical points

Set $y = 20t(t-\sqrt{3})(t+\sqrt{3}) = 0$. The critical points for $t\geq0$ are $t = 0$ and $t=\sqrt{3}$.

Step4: Test the intervals

We test the intervals $[0,\sqrt{3})$ and $(\sqrt{3},\infty)$. For the interval $[0,\sqrt{3})$, let's choose a test - point, say $t = 1$. Then $v(1)=-20\times1^{3}+60\times1=-20 + 60=40>0$. For the interval $(\sqrt{3},\infty)$, let's choose $t = 2$. Then $v(2)=-20\times2^{3}+60\times2=-160 + 120=-40<0$.

Answer:

The particle is moving to the right on the interval $[0,\sqrt{3})$.