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A particle moves along the $x$-axis so that at time $t \geq 0$ its velocity is given by $v(t)=9 t^{2}-18 t$. Determine all intervals when the particle is moving to the right.
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Step 1 ：A particle moves along the x-axis so that at time t ≥ 0 its velocity is given by v(t)=9 t^{2}-18 t. We need to determine all intervals when the particle is moving to the right.

Step 2 ：The particle is moving to the right when its velocity is positive. Therefore, we need to find the intervals of t for which v(t) > 0. This involves solving the inequality 9t^2 - 18t > 0.

Step 3 ：Solving the inequality gives us two intervals: t < 0 and t > 2.

Step 4 ：However, the problem states that t ≥ 0, so we can ignore the interval t < 0.

Step 5 ：Therefore, the particle is moving to the right when t > 2.

Step 6 ：Final Answer: \(\boxed{t > 2}\)