rac{d}{dt}(42 - 52t + 22t^2) = -52 + 44t = 0 \Rightarrow t = rac{13}{11}
["# Understanding the Equation: Solving ( \frac{dr}{dt}(42 - 52t + 22t^2) = -52 + 44t = 0 \Rightarrow t = \frac{13}{11} )", "When solving physics problems involving motion, especially those related to position, velocity, and acceleration, calculus plays a fundamental role. One common challenge encountered is determining the time at which the velocity of an object equals zero. This occurs at a critical moment—such as when an object pauses momentarily before changing direction. In this article, we explore a key mathematical derivation where the derivative of a quadratic position function reveals when the velocity (given by ( \frac{dr}{dt} )) is zero, leading to the solution ( t = \frac{13}{11} ).", "## The Physics and Math Behind Velocity and Acceleration", "Assume we have a position function ( r(t) ) describing the location of an object along a straight line, defined as:", "[\nr(t) = 42 - 52t + 22t^2\n]", "The velocity ( v(t) ) is the first derivative of position with respect to time:", "[\nv(t) = \frac{dr}{dt} = -52 + 44t\n]", "The acceleration ( a(t) ) is the derivative of velocity, or the second derivative of position:", "[\na(t) = \frac{d^2r}{dt^2} = 44\n]", "Since acceleration is constant in this case, velocity changes linearly over time. Setting the velocity equal to zero identifies the exact moment the object momentarily stops:", "[\n\frac{dr}{dt} = -52 + 44t = 0\n]", "## Solving for When Velocity Equals Zero", "To find when this critical moment occurs, solve:", "[\n-52 + 44t = 0\n]", "Add 52 to both sides:", "[\n44t = 52\n]", "Divide both sides by 44:", "[\nt = \frac{52}{44} = \frac{13}{11}\n]", "Thus, the velocity is zero at ( t = \frac{13}{11} ) seconds. This result is significant—whether modeling projectile motion, vehicular motion, or mechanical systems—knowing the precise time when velocity crosses zero enables better analysis of motion dynamics, such as calculating displacement, maximum height, or deceleration phases.", "## Why ( t = \frac{13}{11} ) Matters in Real-World Applications", "- Projectile Motion: At ( t = \frac{13}{11} ), a projectile reaches its peak altitude before descending.\n- Engineering Systems: In mechanical parts or engines, detecting zero velocity helps avoid collisions or optimize stopping times.\n- Mathematical Modeling: This simple quadratic-function derivative exemplifies how calculus simplifies solving dynamic systems in science and engineering.", "## Conclusion", "The equation ( \frac{dr}{dt}(42 - 52t + 22t^2) = -52 + 44t = 0 ) exemplifies a foundational concept: finding when an object's velocity is zero using derivatives. Solving ( -52 + 44t = 0 ) yields ( t = \frac{13}{11} ), a precise moment invaluable in physics and applied mathematics. Understanding this derivation enhances problem-solving skills, enabling clearer interpretations of motion and fostering deeper insight into calculus in real-world contexts.", "---", "Keywords:\n( \frac{d}{dt}(42 - 52t + 22t^2) = -52 + 44t ), velocity = 0, solve ( -52 + 44t = 0 ), t = 13/11, calculus in motion, physics derivatives, acceleration constant, projectile motion, engineering dynamics."]
