Distance And Displacement Worksheet: Answers & Insights
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Distance and displacement are fundamental concepts in physics and everyday life that help us understand motion. While they might seem interchangeable at first glance, there are significant differences that can influence how we perceive and measure movement. This article provides insights into distance and displacement, along with a detailed worksheet that includes answers to some typical questions and problems involving these concepts. 🚀
Understanding Distance and Displacement
What is Distance?
Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. It does not consider the direction of the path taken; it simply measures the total length traveled. For example, if you walk 3 meters to the east and then 4 meters back to the west, the total distance you have walked is:
[ \text{Total Distance} = 3 \text{ m} + 4 \text{ m} = 7 \text{ m} ]
What is Displacement?
On the other hand, displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position. Displacement takes direction into account. Using the same example as above, if you walk 3 meters east and then return 4 meters west, your displacement would be:
[ \text{Displacement} = \text{Final Position} - \text{Initial Position} = -1 \text{ m} \text{ (to the west)} ]
Key Differences between Distance and Displacement
| Feature | Distance | Displacement |
|---|---|---|
| Definition | Total path length traveled | Shortest path between two points |
| Quantity Type | Scalar | Vector |
| Direction | Does not consider direction | Considers direction |
| Positive/Negative | Always positive | Can be positive or negative |
| Calculation | Sum of all distances traveled | Final position - Initial position |
Insights into Distance and Displacement
Understanding the difference between distance and displacement is vital in physics, navigation, and various real-life situations. Here are some key insights:
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Vector vs. Scalar: Recognizing the distinction helps in applying mathematical operations correctly. When dealing with displacement, one must always consider the direction.
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Real-World Applications: In navigation, distance can inform you of how much distance you’ve traveled, while displacement can inform you of your actual position relative to your starting point.
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Impact on Results: In various physics problems, using distance instead of displacement (or vice versa) can lead to incorrect conclusions, particularly in calculating speed, velocity, and acceleration.
Distance and Displacement Worksheet
Now that we've established the foundational concepts, let's apply them with a practical worksheet. Below are several questions with varying difficulty, along with answers provided at the end.
Questions
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Question 1: A car travels 40 km north, then 15 km east. What is the total distance traveled, and what is the displacement?
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Question 2: A person jogs 5 km south and then walks 3 km north. What is their total distance and displacement?
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Question 3: If a cyclist goes around a circular track with a circumference of 200 m, and returns to the starting point, what is the distance and displacement?
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Question 4: A drone flies from point A to point B, a straight-line distance of 100 m, and then moves back to point A. Calculate both the distance and displacement.
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Question 5: A hiker walks 6 km east, then 8 km north. What is the distance traveled and the displacement?
Answers
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Answer 1:
- Total Distance: 40 km + 15 km = 55 km
- Displacement: Use Pythagorean theorem:
[ \text{Displacement} = \sqrt{(40^2 + 15^2)} = \sqrt{1600 + 225} = \sqrt{1825} \approx 42.72 \text{ km} \text{ (at a north-east angle)} ]
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Answer 2:
- Total Distance: 5 km + 3 km = 8 km
- Displacement: 5 km south - 3 km north = 2 km south
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Answer 3:
- Distance: 200 m
- Displacement: 0 m (as the cyclist returns to the starting point)
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Answer 4:
- Total Distance: 100 m + 100 m = 200 m
- Displacement: 0 m (returns to starting point)
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Answer 5:
- Total Distance: 6 km + 8 km = 14 km
- Displacement: Use Pythagorean theorem:
[ \text{Displacement} = \sqrt{(6^2 + 8^2)} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ km} \text{ (at a north-east angle)} ]
Conclusion
In summary, distance and displacement are two distinct concepts essential for understanding motion. The distance represents the total path traveled regardless of direction, while displacement measures the shortest route from the starting point to the end point, including direction. By working through examples and engaging with worksheets, we can solidify our understanding of these fundamental concepts. Always remember: the distinction is crucial in both theoretical and practical applications! 🌍