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Representing Motion Graphically
 Displacement Against Time Graphs (s/t graphs) In this graph: the lines, a, represent stationary bodies the lines, b, represent motion with uniform (constant) velocity This graph represents accelerated motion If the acceleration is uniform then the curve is parabolic since s a t2 The slope of an s/t graph represents velocity Velocity Against Time Graphs (v/t graphs) In this graph: the line, a, represents uniform velocity the lines, b, represent uniform acceleration The lines on this graph represent non-uniform accelerations The slope of a v/t graph represents acceleration v/t graph for a body in free fall Using a v/t Graph to Find Distance Moved (or Displacement) Here, the body moved with constant velocity During the 8s period shown, it is clear that the body moved 8 × 10 = 80m In this case, the body was accelerating, uniformly, so the average speed during the 6s period was 15/2 = 7.5ms-1 therefore, the distance moved was 7.5 × 6 = 45m Notice that, in each case, the arithmetic needed to find the distance moved by the body is equivalent to finding the area between the line and the time axis. area of rectangle = base × height area of rectangle = ½(base × height) In these simple cases, this observation changes nothing about the calculation needed to find the distance moved by the body However, if the shape of the graph is more complicated, as shown here, the only way to find the distance moved might be to draw the graph and measure the area under the curve (or more likely ask a computer to do it for you). The following graph represents a situation in which the sense of the motion of the body changed Combining the areas between the graph and the time axis, we have A1 = +12 A2 = +12 A3 = -1 A4 = -6 This means that the body moved a total distance of 31m. However, because of the change in the sense of the motion at t = 6s, the final displacement of the body at t = 10s, from its position at t = 0, is (24 - 7)m = 17m  in the positive sense.
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