The area can be subsequently determined using the appropriate formula. Once calculated, this area represents the displacement of the object. The widget below computes the area between the line on a velocity-time plot and the axes of the plot. This area is the displacement of the object.
Use the widget to explore or simply to practice a few self-made problems. We Would Like to Suggest Sometimes it isn't enough to just read about it. You have to interact with it! And that's exactly what you do when you use one of The Physics Classroom's Interactives. Question 2 Assume the woman in the graph below walks east and west along the city block she lives on, so the y-axis is in units of city blocks and the x-axis is in units of minutes.
Indicate her position at the following time-points: 0, 1. Question 2 Answer 1. Question 3 Using the displacement vs. Convert the graph from a displacement vs. Draw your answer. Question 3 Answer 12 blocks; see graph below The total distance is found by adding the amount of blocks walked in each portion of the graph. The woman walks 2 blocks out, and then 6 blocks backward, and finally 4 blocks forward, for a total distance of 12 blocks.
The key difference in a distance vs. The total average velocity for the trip is equal to the displacement over the total time. Since the displacement for the trip is zero, the average velocity is zero as well. Question 5 In the displacement vs. Interpret velocity vs. Velocity at a given time can be read directly from the y-axis Acceleration is the slope of a velocity vs. The plot below shows her velocity during her walk. Indicate her velocity and acceleration at each of the following time-points: 0.
At what time does she walk south past her house? The velocity is found by reading the value off the graph, and is undefined if there is a vertical line at that point. The acceleration is given by the slope of the graph, and is undefined if the curve changes slope at that point. Finally, the total distance is given by the area under the curve, and the woman passes her home at the point where the total distance is equal to zero again, which is after 4 minutes areas above the axis are positive, areas below are negative, and at 4 minutes positive and negative areas each equal 1.
The total distance is found by adding the areas, and will thus be 1. The displacement, however, takes into account the direction and thus the second area must be subtracted: 1.
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