Explain and give the answer ---**Question Stem:** A particle moves along a line with position given by s(t) = t^3 - 6t^2 + 9t for t ≥ 0. At what time is the particle at rest? **Mathematical Formula:** s(t) = t^3 - 6t^2 + 9t **Condition:** t ≥ 0 **Question:** At what time is the particle at rest? **Options:** A. t = 1 B. t = 2 C. t = 3 D. t = 4

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In this problem, we have a particle moving along a line with position given by s of t equals t cubed minus 6 t squared plus 9 t, for t greater than or equal to 0. We need to find at what time the particle is at rest. Let's start by understanding what it means for a particle to be at rest. A particle is at rest when its velocity is zero. To find the velocity, we need to take the derivative of the position function. To find when the particle is at rest, we need to find the velocity function by taking the derivative of the position function. The position function is s of t equals t cubed minus 6 t squared plus 9 t. Taking the derivative, we get v of t equals 3 t squared minus 12 t plus 9. The particle is at rest when its velocity is zero. So we need to solve the equation 3 t squared minus 12 t plus 9 equals 0. Looking at the graph of the velocity function, we can see that it crosses the x-axis at two points: t equals 1 and t equals 3. These are the times when the particle is at rest. Now, let's solve the equation 3 t squared minus 12 t plus 9 equals 0. First, we can factor out the common factor 3, giving us 3 times t squared minus 4t plus 3 equals 0. Dividing both sides by 3, we get t squared minus 4t plus 3 equals 0. This quadratic equation can be factored as t minus 1 times t minus 3 equals 0. Therefore, either t equals 1 or t equals 3. Since we're told that t is greater than or equal to 0, both solutions are valid. So the particle is at rest at t equals 1 and t equals 3. Looking at our velocity graph, we can confirm that the velocity is zero at these two points. Let's verify our solutions by substituting t equals 1 and t equals 3 into the velocity function. At t equals 1, v of 1 equals 3 times 1 squared minus 12 times 1 plus 9, which equals 3 minus 12 plus 9, which equals 0. At t equals 3, v of 3 equals 3 times 3 squared minus 12 times 3 plus 9, which equals 27 minus 36 plus 9, which equals 0. So we've confirmed that the particle is at rest at t equals 1 and t equals 3. Looking at the position graph, we can see that the tangent lines at these points are horizontal, indicating zero velocity. Since the question asks for the time when the particle is at rest, and option C is t equals 3, the answer is C. Let's summarize what we've learned. A particle is at rest when its velocity is zero. To find the velocity, we take the derivative of the position function. For our position function s of t equals t cubed minus 6 t squared plus 9 t, the velocity function is v of t equals 3 t squared minus 12 t plus 9. Setting the velocity equal to zero and solving the quadratic equation, we found that t equals 1 or t equals 3. Both are valid solutions since they satisfy the condition t is greater than or equal to 0. Therefore, the particle is at rest at t equals 1 and t equals 3. Since the question asks for the time when the particle is at rest, and option C is t equals 3, the answer is C.