The motion graph shown below was created by a toy train which starts out moving north. The train starts from a position of 2.0m north.
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Answer:
The motion graph provided represents the displacement of a toy train over time. The graph consists of two distinct segments: an initial period of constant velocity followed by a period of rest.
From the given information, we can determine that the train starts from a position of 2.0m north. This means that at t=0 (the beginning of the graph), the train is located 2.0m north of its starting point.
The first segment of the graph shows a straight line with a positive slope, indicating constant velocity. Since the train is moving north, the positive slope suggests that it is moving in the positive direction along the y-axis. The steeper the slope, the greater the velocity.
The second segment of the graph shows a horizontal line, indicating that the train is at rest. During this period, the train does not undergo any displacement and remains stationary.
To determine the total displacement of the train, we need to calculate the area under the graph. In this case, we have two separate areas to consider: one for each segment.
For the first segment, which represents motion, we can calculate the area by finding the area of a triangle. The formula for calculating the area of a triangle is A = 1/2 * base * height. In this case, the base corresponds to the time interval and the height corresponds to the displacement.
Let's assume that each unit on both axes represents 1 second and 1 meter, respectively. From the graph, we can estimate that the time interval for the first segment is approximately 4 seconds and that the displacement is approximately 8 meters (from t=0 to t=4). Therefore, using our formula, we can calculate:
A = 1/2 * 4s * 8m = 16m²
So, during this period of motion, the train has a displacement of 16 meters.
For the second segment, which represents rest, the train does not undergo any displacement. Therefore, the area under the graph is zero.
To calculate the total displacement of the train, we sum up the areas from both segments:
Total displacement = 16m² + 0m² = 16m²
Hence, the total displacement of the toy train is 16 meters.
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