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On a jet-powered car's position-time graph, two points are (0.50 s, 525 m) and (6.4 s, 2000 m). What is the car's average velocity between these points?
A150 m/s
B320 m/s
C250 m/s
D400 m/s
Answer & Solution
Correct answer: C. 250 m/s
1. Average velocity is the slope between the two points: (change in position) / (change in time).
2. Change in position = 2000 m − 525 m = 1475 m. Change in time = 6.4 s − 0.50 s = 5.9 s.
3. Average velocity = 1475 m / 5.9 s = 250 m/s.
4. This matches the text's own cross-check in km/h: "an impressively high land speed (900 km/h..." and 250 m/s x 3.6 = 900 km/h exactly.
5. So the car's average velocity between these two points is 250 m/s.
_Source: OpenStax Physics Ch 2 "Motion in One Dimension", p.73 Worked Example §2.3 Position vs. Time Graphs_
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