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Flip to reveal answersOn a velocity–time graph, what does the area under the line give?
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All 11 Flashcards — Displacement from a velocity–time graph
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Question
On a velocity–time graph, what does the area under the line give?
Answer
The **displacement** — how far the object travels.
Question
On a velocity–time graph, what does the slope give?
Answer
The **acceleration**. (Area = displacement, slope = acceleration — don't swap them.)
Question
What is the given data-booklet formula for displacement from a straight v–t line?
Answer
$s = \dfrac{u + v}{2}\,t$ — average velocity × time (the trapezium area).
Question
What does ½(u + v) represent?
Answer
The **average velocity** — halfway between the start velocity u and the final velocity v.
Question
Area of a triangle under a v–t line (from rest)?
Answer
**½ × base × height** = ½ × time × final velocity.
Question
Area of a rectangle under a flat v–t line?
Answer
**speed × time** — a constant velocity gives a rectangular area.
Question
How do you handle an awkward area under a v–t graph?
Answer
**Split it** into a rectangle + a triangle, work out each, then **add** them.
Question
A v–t line dips below the time axis. What does that area mean?
Answer
**Negative** displacement — the object is moving backwards. Subtract it from the forward area for the net displacement.
Question
A v–t line is flat at 10 m s⁻¹ for 3.0 s. Displacement?
Answer
Rectangle area = 10 × 3.0 = **30 m**.
Question
A v–t line rises from rest to 12 m s⁻¹ over 4.0 s. Displacement?
Answer
Triangle area = ½ × 4.0 × 12 = **24 m**.
Question
Why does the unit of a v–t area come out in metres?
Answer
Height (m s⁻¹) × width (s) = m s⁻¹ × s = **m** — exactly a displacement.
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Topic 1.1 hub
Kinematics
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