The graph shows the speed of a remote controlled toy speed boat on a river. It speeds up for six seconds and then slows down for the same amount of time. The boat's displacement (in metres) can be calculated from the area under this graph. What is it?

Question

The graph shows the speed of a remote controlled toy speed boat on a river. It speeds up for six seconds and then slows down for the same amount of time. The boat's displacement (in metres) can be calculated from the area under this graph. What is it?

BlackTom AI · Accepted Answer

First, I restate the given information to be sure we’re analyzing the same scenario: the boat’s speed increases linearly for six seconds and then decreases at the same rate for the next six seconds, forming a symmetric speed–time shape on the graph. The quantity asked is the displacement, which by definition is the area under a speed–time graph.

Now, analyze the options (or, in this case, the single provided statement):
- The statement says: 'The displacement is calculated as the area of the triangle under the speed-time graph.' This aligns with the fundamental concept that displacement equals the integral (area) under the speed–time curve. Since the speed changes linearly from 0 up to a maximum and back to 0 over the total 12 seconds, the graph forms a triangle with base equal to the total time (12 s) and height equal to the maximum speed.
- Why a triangle? Because a linear increase from 0 to vmax over 6 s creates one right triangle when considering the area from t = 0 to t = 6 s, and the subsequent linear decrease from vmax back to 0 over the next 6 s creates an identical triangle on the remaining interval. The combination yields a single triangular area when viewed from t = 0 to t = 12 s with the baseline along the time axis and the peak at vmax.

If we wanted to be more explicit about the geometry:
- The base of the triangle is the total time of motion (12 s).
- The height is the maximum speed reached (vmax).
- The area (displacement) is (1/2) × base × height = (1/2) × 12 × vmax = 6 × vmax (in metres, since speed is in m/s and time in s).

Thus, the statement identifying the displacement with the area under the speed–time graph being a triangle is conceptually correct for this scenario, and the displacement is computed via that triangular area. If there were other options, we would compare them by applying the same idea that displacement corresponds to the area under the speed–time curve and check for any misinterpretations (e.g., using distance or misidentifying units or shapes).

The graph shows the speed of a remote controlled toy speed boat on a river. It speeds up for six seconds and then slows down for the same amount of time. The boat's displacement (in metres) can be calculated from the area under this graph. What is it?

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