The position of a particle moving along a line is given by \( s(t) = t^3 - 6t^2 + 9t + 2 \), where \( s \) is in meters and \( t \) in seconds. Find the time(s) when the particle is at rest.
["Optimal Moment By Which a Particle Is at Rest: Analyzing Motion Along a Line Using ( s(t) = t^3 - 6t^2 + 9t + 2 )", "Understanding when a particle moves at rest is fundamental in physics and motion analysis. A particle is at rest when its velocity is zero—this means the slope of the position function ( s(t) ) is zero. In this article, we analyze the motion of a particle described by the position function:", "[\ns(t) = t^3 - 6t^2 + 9t + 2\n]", "where ( s ) is measured in meters and ( t ) in seconds. We will determine the time(s) when the particle is at rest by computing the velocity function and solving for when it equals zero.", "---", "### What Does It Mean for a Particle to Be at Rest?", "A particle moving along a line is at rest when its instantaneous velocity is zero. Mathematically, velocity ( v(t) ) is the derivative of position with respect to time:", "[\nv(t) = s'(t)\n]", "Thus, we must compute ( s'(t) ), set it equal to zero, and solve for ( t ).", "---", "### Step 1: Compute the Velocity Function", "Differentiate ( s(t) = t^3 - 6t^2 + 9t + 2 ):", "[\nv(t) = s'(t) = 3t^2 - 12t + 9\n]", "---", "### Step 2: Find When the Particle Is At Rest", "Set velocity equal to zero:", "[\n3t^2 - 12t + 9 = 0\n]", "Divide the entire equation by 3 to simplify:", "[\nt^2 - 4t + 3 = 0\n]", "Factor the quadratic:", "[\n(t - 1)(t - 3) = 0\n]", "Solve for ( t ):", "[\nt = 1 \quad \ ext{or} \quad t = 3\n]", "---", "### Step 3: Interpret the Results", "At ( t = 1 ) second and ( t = 3 ) seconds, the particle’s velocity is zero, meaning it is momentarily at rest. This occurs because the slope of the position curve touches zero at these points—indicating turning points or pauses in motion.", "---", "### The Role of Acceleration in Motion Analysis", "To confirm whether the particle changes direction (and thus truly "comes to rest"), examine the acceleration ( a(t) = v'(t) = 6t - 12 ). At ( t = 1 ), ( a = -6 ) m/s² (negative), and at ( t = 3 ), ( a = 6 ) m/s² (positive), indicating a change in direction at both points, so both times correspond to true stationary pauses.", "---", "### Conclusion", "For the particle moving along a line with position function ( s(t) = t^3 - 6t^2 + 9t + 2 ), the times when the particle is at rest are:", "[\n\boxed{t = 1 \ ext{ second and } t = 3 \ ext{ seconds}}\n]", "These moments are critical for analyzing motion dynamics, velocity changes, and position behavior—key concepts in kinematics and physics education.", "---", "Keywords: particle motion, velocity function, at rest times, ( s(t) ), calculus of position, kinematics, ( v(t) = 0 ), ( t^3 - 6t^2 + 9t + 2 ), time intervals, rest points, acceleration analysis."]
