Worked example: distance and displacement from position-time graphs | AP Physics 1 | Khan Academy
TLDRThis video script delves into the concepts of distance and displacement, using position versus time graphs to illustrate these principles. The instructor guides viewers through two examples involving an iguana and an armadillo, showing how to calculate both displacement and distance. For the iguana, the displacement between zero and five seconds is eight meters to the right, while the distance traveled is also eight meters. If the time frame extended to six seconds, the displacement would be two meters to the right, and the distance would be fourteen meters. The armadillo example demonstrates a displacement of negative six meters (to the left) over 24 seconds, with a total distance traveled of 24 meters. The video emphasizes the importance of direction in displacement and the total path length in distance, providing a clear understanding of these fundamental physics concepts.
Takeaways
- 📐 **Displacement vs Distance**: Displacement is the change in position considering direction, while distance is the total path length traveled without regard to direction.
- ⏱️ **Time Frame Clarity**: It's crucial to focus on the exact time frame specified for calculating displacement and distance.
- 🐊 **Iguana Example**: For the iguana running between 0 and 5 seconds, the displacement is +8 meters, indicating a shift to the right.
- 📏 **Calculating Displacement**: Displacement equals the final position minus the initial position, taking into account the direction of movement.
- 🔢 **Initial and Final Positions**: To find displacement, identify the starting and ending positions within the given time frame.
- 🏃 **Iguana's Path**: The iguana's path from -2 meters to +6 meters over 3 seconds before remaining at +6 meters until the 5-second mark.
- 🛤️ **Distance Calculation**: Distance is the sum of the lengths of the individual legs of the journey, regardless of direction.
- 🕒 **Time Extension Impact**: Extending the time frame from 5 to 6 seconds for the iguana changes the displacement from +8 to +2 meters.
- 🦎 **Armadillo Example**: Over 24 seconds, the armadillo's displacement is -6 meters, indicating a shift to the left.
- 🔄 **Total Path Length**: The armadillo's total distance traveled is 24 meters, which is the sum of all movements during the 24 seconds.
- ➡️ **Directional Significance**: Displacement's negative or positive sign indicates the direction of movement (left or right).
Q & A
What is the difference between distance and displacement?
-Distance is the total length of the path traveled by an object, regardless of direction. Displacement, on the other hand, is the change in position of an object and takes into account the direction of the movement.
How is displacement calculated?
-Displacement is calculated by subtracting the initial position from the final position of an object. It is a vector quantity, which means it has both magnitude and direction.
What is the displacement of the iguana between zero seconds and five seconds according to the graph?
-The displacement of the iguana between zero seconds and five seconds is 8 meters in the positive direction.
How is distance calculated when an object changes direction during its movement?
-Distance is the sum of the lengths of each segment of the path traveled by the object, regardless of the direction of each segment.
What is the total distance traveled by the iguana between zero seconds and five seconds?
-The total distance traveled by the iguana between zero seconds and five seconds is 8 meters.
What would be the displacement of the iguana if the time frame was considered up to six seconds instead of five seconds?
-If the time frame was considered up to six seconds, the displacement of the iguana would be 2 meters in the positive direction.
What is the total distance traveled by the iguana between zero seconds and six seconds?
-The total distance traveled by the iguana between zero seconds and six seconds is 14 meters.
What is the displacement of the armadillo over the first 24 seconds according to the graph?
-The displacement of the armadillo over the first 24 seconds is -6 meters, indicating a shift to the left.
What is the total distance traveled by the armadillo over the first 24 seconds?
-The total distance traveled by the armadillo over the first 24 seconds is 24 meters.
Why is the displacement of the armadillo negative over the first 24 seconds?
-The displacement is negative because the armadillo ends up 6 meters to the left of its starting position, indicating a leftward movement.
How does the direction of movement affect the calculation of displacement?
-The direction of movement affects the calculation of displacement because displacement is a vector quantity. Positive values indicate movement in one direction (usually to the right or forward), while negative values indicate movement in the opposite direction.
Why is it important to consider both distance and displacement when analyzing motion?
-It is important to consider both distance and displacement when analyzing motion because distance provides the total path length traveled, while displacement provides information about the change in position and direction of the object.
Outlines
📈 Understanding Displacement and Distance with Iguana's Example
This paragraph explains the concept of displacement and distance using the example of a 3.2 kilogram iguana running back and forth along the ground. The instructor emphasizes the importance of considering direction when calculating displacement, which is the change in position from the start to the end. The iguana's final position at 5 seconds is 6 meters, and the initial position is -2 meters, resulting in a displacement of +8 meters to the right. The distance traveled, which is the total path length without considering direction, is also 8 meters. The paragraph further explores what would happen if the time frame was extended to 6 seconds, resulting in different displacement and distance values.
🐦 Analyzing Armadillo's Movement in the Desert
The second paragraph discusses the movement of a 2.7 kilogram armadillo rolling in a straight line in the desert. The focus is on calculating the displacement and distance traveled over the first 24 seconds. The armadillo's final position at 24 seconds is 0 meters, and the starting position is 6 meters, leading to a displacement of -6 meters to the left. To find the distance, the armadillo's journey is broken down into segments: it starts at 6 meters, stays there for 8 seconds, then moves to 15 meters by the 16th second, and finally returns to 0 meters. The distance is the sum of the absolute values of these movements, which equals 24 meters. The paragraph highlights the difference between displacement, which considers direction and results in a net shift, and distance, which only considers the length of the path traveled, regardless of direction.
Mindmap
Keywords
💡Displacement
💡Distance
💡Position Versus Time Graph
💡Direction
💡Iguana
💡Armadillo
💡Magnitude
💡Scalar Quantity
💡Vector Quantity
💡Initial Position
💡Final Position
Highlights
The concept of displacement is explained as the change in position with direction considered.
Displacement is calculated as the final position minus the initial position.
The iguana example demonstrates how to calculate displacement over a time period.
Distance is defined as the total path length traveled, regardless of direction.
For the iguana, both displacement and distance over the first three seconds are eight meters.
The difference between displacement and distance is highlighted with a change in time from five to six seconds.
If the time frame is extended to six seconds, the iguana's displacement changes to two meters.
The total distance traveled by the iguana over six seconds is 14 meters.
An armadillo example is used to further illustrate the concepts of displacement and distance.
The armadillo's displacement over 24 seconds is calculated to be negative six meters, indicating a leftward shift.
The armadillo's total distance traveled is 24 meters, despite the displacement being negative six meters.
Distance is the sum of the lengths of the individual paths traveled, without considering direction.
The armadillo's path is broken down into segments to calculate the total distance.
The importance of understanding the difference between displacement (which includes direction) and distance (which does not) is emphasized.
The concepts are applied to real-world examples to help viewers understand the practical applications.
Visual aids such as number lines are used to clarify the direction and magnitude of displacement.
The video concludes by reinforcing that displacement is concerned with direction, while distance is only about the total path length.
Transcripts
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