The graph shows the speed of a remote controlled toy car on a pathway. It speeds up for one and a half seconds and then slows down for the same amount of time. 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

The graph shows the speed of a remote controlled toy car on a pathway. It speeds up for one and a half seconds and then slows down for the same amount of time. 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.

BlackTom AI · Accepted Answer

The task asks for the displacement of the toy car, which corresponds to the area under a velocity–time graph.
First, note the given motion: the car speeds up for 1.5 s and then slows down for the same amount of time, returning to zero velocity at the end. This creates a symmetric triangular velocity–time profile with a base (total time) of 3.0 s and a peak velocity reached at 1.5 s.
The area under a velocity–time graph for a triangular shape is 1/2 × base × height, where the base is the total time interval and the height is the peak velocity reached.
From the graph, the peak velocity (the height of the triangle) is 3 m/s. Therefore, the displacement is:
Displacement = (1/2) × (base = 3.0 s) × (height = 3.0 m/s) = 0.5 × 3.0 × 3.0 = 4.5 m.
Since the answer must be given to 3 significant figures, we report 4.50 m.

Why the other potential interpretations would be wrong:
- If one thought the area was simply base × height without the 1/2 factor, you’d get 9.0 m, which is twice the correct value for a triangular area.
- If you assumed a different peak velocity or base time not matching the graph’s symmetry, the product base × height would yield a different result, contradicting the given 1.5 s up and 1.5 s down pattern.
- If you looked for a rectangular area or a sum of rectangular slices, you’d misinterpret the triangular shape; the velocity increases linearly to a maximum and then decreases linearly, which forms a triangle, not a rectangle.

In summary, applying the correct triangular area formula with base 3.0 s and peak velocity 3.0 m/s yields the displacement of 4.50 m.

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