Net Displacement with Pythagoras

Net Displacement और Pythagoras

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Net Displacement with Pythagoras

  • Direction Sense
  • Net Displacement with Pythagoras
Hello दोस्तों! MeraExam की एक और class में आपका स्वागत है। आज की class में समझेंगे — Net Displacement और Pythagoras। बिलकुल zero से, एकदम आसान भाषा में। चलिए शुरू करते हैं!
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Learning Objective

Find the shortest distance and final direction from the start by tracking net East-West and North-South movement.

🎯 Learning Objective

Find the shortest distance and final direction from the start by tracking net East-West and North-South movement.

💡 Concept

  • Split every move into East-West (horizontal) and North-South (vertical)
  • Add up: net East-West and net North-South separately (opposite moves cancel)
  • Shortest (straight-line) distance = √(horizontal² + vertical²)
  • Final direction is the corner between the two net directions (e.g. North + East = North-East)
  • Watch for standard triples: 3-4-5, 5-12-13, 8-15-17 — they save calculation

🧮 Key Formulas

Shortest distance = √(net East-West² + net North-South²)

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Triples: 3-4-5, 5-12-13, 8-15-17

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Opposite moves cancel out

✏️ Easy Example

Q. Ravi walks 3 km towards the East, then 4 km towards the North. How far is he from the starting point?

  1. Net East = 3 km, net North = 4 km
  2. Distance = √(3² + 4²) = √(9 + 16) = √25

Answer: 5 km

🇮🇳 Real-Life Example

Walking blocks in a planned city like Chandigarh — 3 blocks across, 4 blocks up — the crow-flight distance home is this same right-triangle shortcut.

📝 Exam-Level Example

Q. Aman starts from home, walks 10 m North, turns East and walks 6 m, then turns South and walks 4 m. How far and in which direction is he from home?

  1. Net North-South: 10 North − 4 South = 6 North
  2. Net East-West: 6 East
  3. Distance = √(6² + 6²) = √72 = 6√2 ≈ 8.49 m; direction is North-East

Answer: 6√2 m (≈ 8.49 m), North-East

📝 Exam-Level Example

Q. A person walks 5 km towards the South, then 12 km towards the East. Find the shortest distance and direction from the start.

  1. Net South = 5 km, net East = 12 km
  2. Distance = √(5² + 12²) = √(25 + 144) = √169 = 13
  3. He is to the South and East of start → South-East

Answer: 13 km, South-East

📝 Exam-Level Example

Q. Rohan drives 8 km West, then 6 km North, then 8 km East. Where is he now relative to his start?

  1. Net East-West: 8 West − 8 East = 0 (they cancel)
  2. Net North-South: 6 North
  3. So he is simply 6 km due North of the start

Answer: 6 km towards the North

🪄 Memory Trick

Draw the path on a rough grid as you read — a quick sketch instantly shows which moves cancel and which corner the answer sits in.

⚠️ Common Mistakes

  • ❌ Adding all distances instead of taking net (opposite directions must cancel)
  • ❌ Forgetting to take the square root at the end
  • ❌ Naming the direction backwards — state it from the START to the person

🏆 Exam Tips

  • ✅ Memorise the 3 common triples so you skip long square-root work
  • ✅ Keep East-West and North-South totals in two separate columns

📌 Summary

  • Split moves into E-W and N-S, then net them
  • Shortest distance = √(net horizontal² + net vertical²)
  • Direction = corner of the two net directions
  • Use triples 3-4-5, 5-12-13, 8-15-17 to save time