Answer Key Physics Distance And Displacement Worksheet Answers

Answer Key: Physics Distance and Displacement Worksheet Answers

This comprehensive guide provides answers and explanations for a typical physics worksheet covering distance and displacement. Understanding the difference between these two fundamental concepts is crucial in mastering kinematics. We'll delve into various examples, clarifying the calculations and providing insights into the underlying principles. This resource serves as a valuable tool for students, teachers, and anyone seeking a deeper understanding of distance and displacement in physics.

Understanding Distance and Displacement: Key Differences

Before diving into the answers, let's solidify our understanding of the core concepts:

Distance: This is a scalar quantity, meaning it only has magnitude (size). It represents the total length of the path traveled by an object. Think of it as the odometer reading in your car – it keeps accumulating the total miles driven, regardless of direction.

Displacement: This is a vector quantity, possessing both magnitude and direction. It represents the shortest distance between the starting point and the ending point of an object's motion. It's a straight line connecting the initial and final positions.

Example: Imagine walking 5 meters east, then 3 meters north.

• Distance: The total distance traveled is 5m + 3m = 8m.
• Displacement: The displacement is the straight-line distance from your starting point to your ending point. Using the Pythagorean theorem (a² + b² = c²), the displacement is √(5² + 3²) ≈ 5.83 meters, in a direction northeast (the exact angle can be calculated using trigonometry).

Worksheet Problem Examples and Solutions

Let's tackle some common problems found in distance and displacement worksheets. We'll break down each problem step-by-step, explaining the reasoning and the calculations involved.

Problem 1: A car travels 10 km east, then 5 km south. Find the total distance and the displacement.

Solution:

• Distance: The total distance is simply the sum of the individual distances: 10 km + 5 km = 15 km.

• Displacement: We need to use the Pythagorean theorem. The eastward displacement is 10 km, and the southward displacement is -5 km (negative because it's in the opposite direction of the positive y-axis). The magnitude of the displacement is √(10² + (-5)²) = √125 ≈ 11.18 km. The direction can be found using trigonometry: tan θ = (-5/10), which gives an angle of approximately -26.6° south of east.

Problem 2: A jogger runs 200 meters north, then turns around and runs 100 meters south. Calculate the distance and displacement.

Solution:

• Distance: The total distance covered is 200 m + 100 m = 300 m.

• Displacement: The displacement is the difference between the final and initial positions. Since the jogger ended up 100 meters north of the starting point, the displacement is 100 meters north.

Problem 3: A bird flies 50 meters due west, then 30 meters due north, and finally 20 meters due east. Determine the total distance and the resultant displacement.

Solution:

• Distance: The total distance is 50 m + 30 m + 20 m = 100 m.

• Displacement: We need to resolve the individual displacements into their x and y components.

  • Westward displacement: -50 meters (negative x-direction)
  • Northward displacement: 30 meters (positive y-direction)
  • Eastward displacement: 20 meters (positive x-direction)

  The net x-displacement is -50 m + 20 m = -30 m. The net y-displacement is 30 m.

  Using the Pythagorean theorem: √((-30)² + 30²) = √1800 ≈ 42.43 meters. The direction is found using trigonometry: tan θ = (30/-30) = -1, indicating an angle of -45° or 45° Northwest of the starting point.

Problem 4: A marble rolls 15 cm east, then 20 cm north, and then 25 cm west. What is the distance and displacement of the marble?

Solution:

• Distance: The total distance is 15 cm + 20 cm + 25 cm = 60 cm.

• Displacement:

  • Eastward displacement: 15 cm
  • Northward displacement: 20 cm
  • Westward displacement: -25 cm (negative x-direction)

  Net x-displacement: 15 cm - 25 cm = -10 cm Net y-displacement: 20 cm

  Magnitude of displacement: √((-10)² + 20²) = √500 ≈ 22.36 cm Direction: arctan(20/-10) ≈ -63.4° (63.4° North of West)

Problem 5 (More Challenging): A hiker walks 2 km north, then 3 km east, then 1 km south. What is the total distance and displacement?

Solution:

• Distance: 2 km + 3 km + 1 km = 6 km

• Displacement:

  • Northward displacement: 2 km -1 km = 1 km
  • Eastward displacement: 3 km

  Magnitude of displacement: √(1² + 3²) = √10 ≈ 3.16 km Direction: arctan(1/3) ≈ 18.4° North of East

Advanced Concepts and Considerations

These problems illustrate fundamental applications of distance and displacement. However, more complex scenarios might involve:

• Curvilinear motion: Instead of straight lines, objects might move along curves. Calculating displacement in such cases requires integrating vector components along the path.

• Three-dimensional motion: Problems could involve movement in three dimensions (x, y, and z axes), requiring vector addition in three-dimensional space.

• Time-dependent motion: The speed or direction of an object might change over time, requiring calculus techniques (integration and differentiation) to determine distance and displacement.

Tips for Solving Distance and Displacement Problems

• Draw diagrams: Visual representation helps significantly in understanding the problem and applying the appropriate formulas.

• Break down complex problems: Divide complex movements into smaller, simpler segments and analyze them individually.

• Use vector notation: Clearly indicate the direction of displacements using plus and minus signs or vector notation (e.g., d = 3i + 4j).

• Master the Pythagorean theorem and trigonometry: These are essential tools for calculating the magnitude and direction of displacements.

• Practice regularly: Solving various types of problems enhances your understanding and problem-solving skills.

Conclusion

Understanding the distinction between distance and displacement is foundational to physics. By mastering the concepts and techniques presented here, you'll be well-equipped to tackle a wide range of kinematics problems. Remember to practice diligently and always break down complex problems into manageable steps. This detailed guide, with its comprehensive solutions and explanations, serves as a valuable resource for solidifying your understanding of distance and displacement in physics. Continuous practice and a thorough understanding of vectors are key to success in this area of physics.