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From 0 to 30 s, a car's velocity vs. time graph is a rectangle of height 20 m/s topped by a triangle of height 140 m/s, both spanning the same 30 s base. What is the car's net displacement over the 30 s?
A600 m
B2,100 m
C2,700 m
D4,200 m
Answer & Solution
Correct answer: C. 2,700 m
1. Displacement equals the area under a velocity vs. time graph.
2. Rectangle area = 20 m/s x 30 s = 600 m.
3. Triangle area = 0.5 x 30 s x 140 m/s = 2,100 m.
4. Total displacement = 600 m + 2,100 m = 2,700 m.
_Source: OpenStax Physics (CC BY 4.0), Ch 2 "Motion in One Dimension", section 2.4 Velocity vs. Time Graphs_
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