Question
Physics 211 - This is one more question on the material covered in the last four questions, starting with the June 13, 1997 question. This question is a variation of the June 27, 1997 question. It deals with graphs of displacement vs. time rather than distance travelled vs. time. (Graphs.)
Suppose a hiker leaves her camp at time t = 0 in the morning and returns to her camp two hours later. Various hypothetical graphs of her displacement from the camp, Dx, vs. time elapsed, t, for the hike are shown above.
Which graph (or graphs) could represent the displacement of the hiker vs. time elapsed? Give reasons, both for the possible and the impossible graphs.
Answer
Graph (d) is the only one that could represent the displacement of the hiker vs. the time elapsed. The four graphs will now be discussed in alphabetical sequence.
Graph (a). Graph (a) shows a displacement Dx that starts with a non-zero value. This is impossible because the displacement is necessarily zero at the start of the motion, which is at t = 0. The fact that the displacement takes on negative values is not a problem. This would happen if the hiker were at a position whose x-value is less than the initial x-value.
Graph (b). Graph (b) is impossible for the same reason as graph (a).
Graph (c). This graph is impossible because it ends up with a non-zero value of the displacement Dx. The displacement should be zero whenever the hiker is at the camp, i.e., at the start and at the finish of the hike.
Graph (d). This graph is possible because the displacement Dx is zero both at the start and at the finish. In between, the displacement is positive at all times. This means the hiker is in one direction away from the camp during her hike and does not pass the camp at any time during the hike.
However, graph (d) is a rather peculiar displacement vs. time graph for a hiker. This becomes apparent when one thinks about the hiker's velocity during her hike. The hiker's velocity will be the subject of next week's question.