Use quadratic formula: \( t = \frac{-32 \pm \sqrt{32^2 + 4 \times 4.9 \times 20}}{2 \times 4.9} \)

["# Use the Quadratic Formula to Solve Motion Problems: A Step-by-Step Guide", "When solving physics problems involving motion under constant acceleration, the quadratic formula is an essential tool. One common application involves determining the time ( t ) it takes for an object to reach a certain position, derived from key kinematic equations. This article explains how to use the quadratic formula—specifically applied to the equation ( t = \frac{-32 \pm \sqrt{32^2 + 4 \ imes 4.9 \ imes 20}}{2 \ imes 4.9} )—to solve quadratic problems efficiently.", "---", "## Understanding the Context: Motion with Constant Acceleration", "In physics, motion with constant acceleration follows the equation:", "[\ns = ut + \frac{1}{2}at^2\n]", "where:\n- ( s ) = displacement\n- ( u ) = initial velocity\n- ( a ) = acceleration\n- ( t ) = time", "Rearranging into standard quadratic form gives:", "[\nat^2 + ut + s = 0\n]", "Depending on given variables, this often leads to a quadratic equation where the quadratic formula solves for time:", "[\nt = \frac{-u \pm \sqrt{u^2 - 4as}}{2a}\n]", "---", "## Analyzing the Given Quadratic Equation", "Consider this specific instance:", "[\nt = \frac{-32 \pm \sqrt{32^2 + 4 \ imes 4.9 \ imes 20}}{2 \ imes 4.9}\n]", "This resembles a rearranged quadratic equation ( at^2 + bt + c = 0 ), where:\n- ( a = 4.9 ) (half acceleration or coefficient of ( t^2 ))\n- ( b = -32 ) (coefficient of ( t ))\n- ( c = 20 ) (the displacement or constant term)", "Although not the standard form, such forms arise when solving motion problems with specific initial conditions and constants.", "---", "## Step-by-Step Calculation Using the Quadratic Formula", "### Step 1: Identify coefficients\nFrom the formula:", "- ( a = 4.9 )\n- ( b = -32 )\n- ( c = 20 )", "Plug into the quadratic formula:", "[\nt = \frac{-(-32) \pm \sqrt{(-32)^2 + 4 \ imes 4.9 \ imes 20}}{2 \ imes 4.9}\n]", "### Step 2: Simplify the discriminant", "Compute the discriminant:", "[\n\Delta = 32^2 + 4 \ imes 4.9 \ imes 20 = 1024 + 392 = 1416\n]", "### Step 3: Apply the quadratic formula", "[\nt = \frac{32 \pm \sqrt{1416}}{9.8}\n]", "Simplify ( \sqrt{1416} ):\nSince ( \sqrt{1416} \approx 37.64 ), we get:", "[\nt = \frac{32 \pm 37.64}{9.8}\n]", "---", "## Step 4: Compute the two possible solutions", "- First root:\n[\nt_1 = \frac{32 + 37.64}{9.8} = \frac{69.64}{9.8} \approx 7.1 \ ext{ seconds}\n]", "- Second root:\n[\nt_2 = \frac{32 - 37.64}{9.8} = \frac{-5.64}{9.8} \approx -0.576 \ ext{ seconds}\n]", "Since time cannot be negative in this physics context, only the positive solution is physically meaningful.", "---", "## Why This Formula Matters in Physics and Engineering", "Using the quadratic formula bridges abstract mathematics and real-world motion analysis. Whether calculating how long it takes for a ball to reach the ground, a car to stop under braking, or launch trajectory timing, solving quadratic equations accurately ensures safe, precise predictions.", "---", "## Final Thoughts", "Mastering quadratic formulas like this empowers students and professionals to solve motion problems confidently. Remember to:\n- Rearrange kinematic equations into standard quadratic form\n- Identify coefficients correctly\n- Calculate the discriminant carefully\n- Select the physically valid time solution", "Next time you encounter a quadratic in physics, treat it as a key to unlocking time, distance, and velocity relationships—crafted clearly by the power of algebra.", "---", "Keywords:\nquadratic formula, motion under acceleration, kinematics, quadratic equation physics, time calculation, physics formulas, solving motion problems, algebra in physics", "Meta Description:\nLearn how to use the quadratic formula with ( t = \frac{-32 \pm \sqrt{32^2 + 4 \ imes 4.9 \ imes 20}}{2 \ imes 4.9} ) to solve motion problems involving time and displacement. Step-by-step guide with physics applications."]