Question
Please refer back to the June 13, 1997 Question. It is available in the Question Archive. (Analytical Geometry. Graphs.)
Draw graphs of (a) position x vs. time t, (b) the corresponding displacement D>x vs. time t, and (c) the corresponding distance s travelled vs. time t for motions from P1 to P2 and back to P1 and for motions from P2 to P1 and back to P2. Do this for an x-axis pointing to the right as in the diagram above and for an x-axis pointing in the opposite direction. That is a total of 12 graphs.
Take x1 = -10 m and x2 = 4 m when the x-axis points to the right and x1 = 10 m and x2 = -4 m when the x-axis points to the left.
Answers
x-axis pointing to the right, motion from P1 to P2 to P1 (first right, then left)
The three graphs of x vs. t, Dx vs. t, and s vs. t are shown below. The time at the start of the motion has been chosen as t = 0.
The first graph shows x vs. t. Note that the particle is not at rest when it is at x = -10 m at t = 0, as indicated by the non-zero slope of the graph at this point. However, the particle gradually slows down to a stop before turning around at x = 4 m, as indicated by the rounded curve at this point. After this point, the graph returns to x = -10 m.
The Dx vs. t graph is identical in shape to the x vs. t graph, but shifted upward because the Dx vs. t graph must start with Dx = 0 at t = 0.
Similarly, the s vs. t graph must start with s = 0 at t = 0. However, a distance travelled can never decrease. Thus, the s vs. t graph keeps on increasing after passing the point at which the particle turns around.
x-axis pointing to the right, motion from P2 to P1 to P2 (first left, then right)
The following three graphs describe this case.
If the motion is of the same type as before, except with a different starting point and in the opposite direction, the x vs. t graph now starts at x = 4 m, drops to x = -10 m, and rises again to x = 4 m.
The displacement vs. time graph is again just the position vs. time graph but shifted upward so that the displacement starts at 0 at t = 0. The distance travelled vs. time graph is exactly the same as before. Distance travelled keeps on increasing whether the motion is to the right or to the left.
x-axis pointing to the left, motion from P1 to P2 to P1 (first right, then left)
The following three graphs describe this case.
The x vs. t graph differs from the previous x vs. t graph only in so far as now the motion is from 10 m to -4 m to 10 m, rather than from 4 m to -10 m to 4 m. Thus, the graph is merely shifted upward by 6 m.
The displacement and distance travelled graphs are identical to the previous case.
x-axis pointing to the left, motion from P2 to P1 to P2 (first left, then right)
The following three graphs describe this case.
The x vs. t graph is very similar to the x vs. t graph in the first case, but now x goes from -4 m to 10 m to -4 m, rather than from -10 m to 4 m to -10 m. Thus, the graph is merely shifted upward by 6 m.
The displacement and distanced travelled vs. time graphs are identical to those in the first case.