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A robin is flying around and its velocity v as a function of time t is given in the graph below where rightwards is the positive velocity direction. What is the robin's displacement Î""x from t = 0s to t = 1.5s?

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Final answer:

The robin's displacement from t = 0s to t = 1.5s can be calculated by finding the area under the velocity-time graph, which represents the displacement. The exact method depends on whether the velocity is constant or changing, and thus the shape of the area - rectangle, trapezoid, or triangle.

Step-by-step explanation:

To determine the robin's displacement (Δx) from t = 0s to t = 1.5s using the velocity-time graph information provided, we need to know the area under the velocity-time curve during this time interval. The graph itself wasn't provided, but assuming a typical straight-line velocity-time graph, the area under the graph from t = 0 to t = 1.5s represents the displacement (since displacement is the integral of velocity with respect to time).

If the velocity-time graph shows a constant velocity, the area would be a rectangle, and the displacement is simply v × t. If the graph shows varying velocity, the area might be a trapezoid or triangle depending on the shape of the line. In this case, we would calculate the area of the geometric shape(s) formed by the graph line from t = 0 to t = 1.5s. For example, for a triangle, the area and thus the displacement would be ½ × base × height, where base is the time interval and the height is the velocity at t = 1.5s.

To find the exact displacement, detailed information about the graph is needed which includes the specific velocities at given times. Without this information, only a general method can be described as done above. However, the student should look at the area under the velocity-time graph in their question to calculate the robin's displacement.

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