The Speeding Ticket Graph
A student records the distance a remote-control car travels over time. At t = 0 s the car is at 0 m. At t = 4 s it is at 12 m, at t = 8 s it is at 24 m, and at t = 12 s it is at 36 m. The student claims, "The graph is a straight line, so the car is speeding up." A classmate disagrees and says a straight line means the speed is constant. The teacher asks the pair to settle the dispute with evidence, then adds a wrinkle: from t = 12 s to t = 16 s the car actually travels only 6 m because it hits a bump. The teacher wants the average speed over the full 16 seconds compared with the speed during the first 12 seconds.
1. Who is correct about the first 12 seconds, and what evidence from the numbers settles it?
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The classmate is correct. A straight-line distance-time graph means equal distances in equal times. Check: 12 m in 4 s, 24 m in 8 s, 36 m in 12 s. Each interval adds 12 m for 4 s, so the unit rate is 12 ÷ 4 = 3 m/s every time. Constant rate of change on a distance-time graph = constant speed. The student confused "line going up" with "steepening." A line that curves upward would mean speeding up; a straight line does not. Evidence: the slope between every consecutive pair of points is 3 m/s.
2. Compute the average speed over all 16 seconds and explain how it differs from the first-12-second speed.
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Total distance = 36 m + 6 m = 42 m. Total time = 16 s. Average speed = 42 ÷ 16 = 2.625 m/s. The first-12-second speed was 3 m/s. The average is lower because the bump interval had a rate of 6 ÷ 4 = 1.5 m/s, which drags the mean down. Important distinction: average speed over the whole trip is not the same as the speed at any moment, and it is not the average of 3 and 1.5 (that would be 2.25). You must weight by time: (3×12 + 1.5×4) ÷ 16 = (36 + 6) ÷ 16 = 2.625 m/s. This shows why total-distance-over-total-time is the reliable method.