A particle moves in a straight line with its position given by \( s(t) = 2t^3 - 5t^2 + 4t \). Find its velocity at \( t = 3 \) seconds.
["Understanding Particle Motion: Finding Velocity From Position Function", "When studying motion in physics, one of the fundamental concepts is determining a particle’s velocity at any given time. If a particle moves along a straight line with its position defined by a mathematical function, velocity is calculated as the derivative of position with respect to time. This article explains how to find the velocity of a particle whose position is given by ( s(t) = 2t^3 - 5t^2 + 4t ) at ( t = 3 ) seconds.", "### How Velocity is Derived from Position", "Velocity describes the rate of change of position over time. Mathematically, velocity ( v(t) ) is the first derivative of the position function ( s(t) ):", "[\nv(t) = \frac{ds}{dt}\n]", "Given the position function:\n[\ns(t) = 2t^3 - 5t^2 + 4t\n]", "Let’s compute the derivative step by step.", "### Step-by-Step Derivative Calculation", "Differentiate each term of ( s(t) ):", "- Derivative of ( 2t^3 ) is ( 2 \cdot 3t^{3-1} = 6t^2 )\n- Derivative of ( -5t^2 ) is ( -5 \cdot 2t^{2-1} = -10t )\n- Derivative of ( 4t ) is ( 4 \cdot 1t^{1-1} = 4 )", "Putting it all together:", "[\nv(t) = \frac{ds}{dt} = 6t^2 - 10t + 4\n]", "### Evaluating Velocity at ( t = 3 ) seconds", "Now substitute ( t = 3 ) into the velocity function:", "[\nv(3) = 6(3)^2 - 10(3) + 4\n]", "Calculate each term:", "- ( 6 \cdot 9 = 54 )\n- ( -10 \cdot 3 = -30 )\n- Constant term is ( 4 )", "Add them:", "[\nv(3) = 54 - 30 + 4 = 28\n]", "### Conclusion", "At ( t = 3 ) seconds, the particle’s velocity along the straight path is 28 meters per second in the direction of motion (positive following the defined coordinate system). This result illustrates how calculus connects position and motion, enabling precise predictions of particle behavior.", "For students and learners, mastering this derivation process is key to understanding kinetic motion and how to analyze real-world dynamic systems using derivatives.", "---", "Keywords: particle motion, position function ( s(t) ), velocity, calculus, ( v(t) = ds/dt ), derivative of ( 2t^3 - 5t^2 + 4t ), ( v(3) ), 3 seconds, physics, kinematics.", "Meta Title: Find the velocity of a particle at t = 3 seconds from s(t) = 2t³ - 5t² + 4t — step-by-step solution\nMeta Description: Learn how to calculate particle velocity using derivatives. Find velocity at t = 3 seconds for s(t) = 2t³ - 5t² + 4t with detailed calculation."]
