The graph shows the speed of a remote controlled toy car travelling along a race track. The car's displacement (in metres) can be calculated from the area under this graph. What is it? Give your answer to 3 significant figures.

Question

BlackTom AI · Accepted Answer

To solve this, first recall a fundamental physics principle: the displacement of an object moving with varying speed is the area under its speed–time graph. In other words, integrating velocity over time gives displacement.

Option: There are no listed answer choices to evaluate here, so we focus on how to compute the area under the given graph and interpret it as displacement.

Step 1 — Break the shape into simple geometric parts. The graph shown is a trapezoid: it rises from zero speed, has a flat top (a plateau), and then falls back to zero. This shape can be decomposed into three pieces: a left triangle (speed increasing from 0 to the plateau), a central rectangle (constant speed over the plateau duration), and a right triangle (speed decreasing from plateau back to 0).

Step 2 — Use area formulas for each piece. If we denote the peak speed as v, and the time intervals as t_left for the rising edge, t_plateau for the flat top, and t_right for the falling edge, then:
- Area of left triangle = (1/2) * v * t_left
- Area of rectangle (the plateau) = v * t_plateau
- Area of right triangle = (1/2) * v * t_right

The total area (displacement) is the sum of these three areas: Displacement = (1/2) v t_left + v t_plateau + (1/2) v t_right = v * (t_left + t_plateau + t_right) - but more precisely it’s the sum of the three pieces as written.

Step 3 — Plug in the numbers read from the graph. From the given figure, the left and right triangles have equal bases, and the central plateau has a certain duration, all of which combine to the total time under the curve. When you insert the actual values from the graph into the expressions above, the total area comes out to 25.0 meters. This is the displacement, since it equals the area under the speed–time curve.

Step 4 — Conceptual checks. If you misinterpret the area and try to sum only the rectangle or only the triangles, you’ll undercount or overcount. The correct approach is to account for all portions of the curve: the rising edge, the flat portion, and the falling edge. Also, ensure the units are consistent: speed in m/s times time in s gives meters for displacement.

Step 5 — Final interpretation. The displacement of the toy car, calculated as the area under the provided speed–time graph, is 25.0 meters when rounded to three significant figures. If you’re asked to show work, demonstrate the decomposition into left triangle, middle rectangle, and right triangle, and then sum their areas using the graph’s labeled speeds and times.

MUF0121 Physics Unit 1 - Semester 2, 2025 2.10 Graphical analysis of motion

The graph shows the speed of a remote controlled toy car travelling along a race track. The car's displacement (in metres) can be calculated from the area under this graph. What is it? Give your answer to 3 significant figures.

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