A train travels 120 km at a speed of \(v\) km/h and then 180 km at \(v + 20\) km/h. If the total time for the journey is 5 hours, find \(v\).

["Find Speed (v) in a Two-Part Train Journey: Detailed Breakdown & Solution", "Trains are a popular mode of transit, but understanding their speed and timing can be puzzling. In this article, we solve a classic problem: a train travels 120 km at speed (v) km/h, followed by 180 km at (v + 20) km/h, with a total journey time of 5 hours. If you’re a student of physics, algebra, or operations research, this article will clarify how to model distance, speed, and time relationships—and find the unknown speed (v).", "---", "### Problem Statement (Recap)", "A train:\n- Travels 120 km at constant speed (v) km/h\n- Then travels 180 km at speed (v + 20) km/h\n- Total trip duration: 5 hours", "Find the value of (v).", "---", "### Understanding the Relationship: Distance = Speed × Time", "The core of this problem lies in the relationship:\n[ \ ext{Time} = \frac{\ ext{Distance}}{\ ext{Speed}} ]", "Let’s compute the travel time for each segment.", "1. First segment:\n Distance = 120 km, Speed = (v) km/h\n Time = ( \frac{120}{v} ) hours", "2. Second segment:\n Distance = 180 km, Speed = (v + 20) km/h\n Time = ( \frac{180}{v + 20} ) hours", "The total time is the sum:\n[\n\frac{120}{v} + \frac{180}{v + 20} = 5\n]", "---", "### Step-by-Step Algebraic Solution", "We solve the equation:\n[\n\frac{120}{v} + \frac{180}{v + 20} = 5\n]", "#### Step 1: Eliminate denominators\nMultiply both sides by (v(v + 20)), the least common denominator:\n[\n120(v + 20) + 180v = 5v(v + 20)\n]", "#### Step 2: Expand both sides\nLeft side:\n[\n120v + 2400 + 180v = 300v + 2400\n]", "Right side:\n[\n5v^2 + 100v\n]", "Now the equation is:\n[\n300v + 2400 = 5v^2 + 100v\n]", "#### Step 3: Bring all terms to one side\nMove all terms to the right to form a quadratic:\n[\n0 = 5v^2 + 100v - 300v - 2400\n]\n[\n5v^2 - 200v - 2400 = 0\n]", "#### Step 4: Simplify the quadratic\nDivide entire equation by 5:\n[\nv^2 - 40v - 480 = 0\n]", "---", "### Step 5: Solve the quadratic equation", "Use the quadratic formula:\n[\nv = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]\nwhere (a = 1), (b = -40), (c = -480)", "[\nv = \frac{40 \pm \sqrt{(-40)^2 - 4(1)(-480)}}{2}\n]\n[\nv = \frac{40 \pm \sqrt{1600 + 1920}}{2}\n]\n[\nv = \frac{40 \pm \sqrt{3520}}{2}\n]", "Simplify (\sqrt{3520}):\nNote that (3520 = 16 \ imes 220 = 16 \ imes 4 \ imes 55 = 64 \ imes 55), so\n[\n\sqrt{3520} = \sqrt{64 \ imes 55} = 8\sqrt{55}\n]", "So:\n[\nv = \frac{40 \pm 8\sqrt{55}}{2} = 20 \pm 4\sqrt{55}\n]", "Now evaluate the positive root:\n(\sqrt{55} \approx 7.416), so\n(4\sqrt{55} \approx 29.664), giving:\n(v \approx 20 + 29.664 = 49.664)", "The other root:\n(v = 20 - 29.664 = -9.664) (invalid, speed can’t be negative)", "Thus,\n[\nv = 20 + 4\sqrt{55}\n]", "---", "### Step 6: Verify the solution (approximate)", "Let’s test (v \approx 49.66)", "- First leg: (120 / 49.66 \approx 2.417) hours\n- Second leg: (v + 20 \approx 69.66), so (180 / 69.66 \approx 2.583) hours\n- Total: (2.417 + 2.583 = 5.000) hours ✅", "Matches the given total time.", "---", "### Final Thoughts: Why This Matters", "This problem blends algebra with real-world application—useful in scheduling, transport logistics, and travel planning. Understanding how to translate word problems into equations is a critical skill in engineering, data science, and everyday decision-making.", "---", "### SEO Keywords for the Article", "- Solve quadratic equations in real-world problems\n- Train journey speed calculation\n- Distance, speed, time problem\n- Algebraic solution for time and distance\n- Find speed given total travel time\n- Train speed problem with two speeds\n- How to calculate average speed in multi-segment trips\n- Applying ( t = d/s ) in practical scenarios", "---", "### Summary", "We began with a two-part train journey, modeled time using ( t = d/s ), set up a rational equation, and reduced it to a solvable quadratic. The positive root gave (v = 20 + 4\sqrt{55} \approx 49.66) km/h. This systematic approach helps anyone tackle similar problems with confidence.", "---", "If you found this breakdown helpful, consider sharing it with others learning algebra or transport physics—because mastering time-speed-distance relationships starts with practice and clear reasoning!", "---", "Author: EduMath Insights | Updated April 2025\nKeywords: train speed, distance-time calculations, quadratic equation, algebra problem, average speed, transport math"]