AP Physics 1 · Unit 1 · Topic 4 of 8 · 6 min lesson

Velocity–Time Graphs

A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.

Quiz this topic · 5 questions Watch the review video ↗

The big idea

On a velocity–time graph, the height is the velocity, the slope is the acceleration, and the area between the graph and the time axis is the displacement. Area below the axis counts as negative.

1. Height, slope, and area

A: speeding up. B: constant velocity. C: slowing down, stops and turns around at t = 7 s, then speeds up in the − direction.

Heightvelocity (sign = direction)

|height|speed

Slope = Δv ÷ Δtacceleration

Area to the time axisdisplacement

Crosses the time axisturns around

2. Speeding up or slowing down?

Look at whether the graph is heading away from or toward the time axis:

Moving away from the axisSpeeding up (v and a have the same sign)

Moving toward the axisSlowing down (v and a have opposite signs)

In segment C the slope never changes, so the acceleration is −3 m/s² the whole time. Before t = 7 s the velocity is positive, so the object slows down. After t = 7 s the velocity is negative too, so the same acceleration now speeds it up in the − direction.

The graph shows velocity, not position. To know where the object is, you also need its starting position.

3. Finding displacement from area

Rectangle: v × Δtconstant velocity

Triangle: ½ × base × heightvelocity changing steadily from 0

Area below the axisnegative displacement

4. Example, then try it

Example

Use the graph below. Find the acceleration during A and the displacement during B.

A: a = Δv ÷ Δt = (6 − 0) m/s ÷ 2 s = +3 m/s²

B: rectangle = (6 m/s)(5 s − 2 s) = +18 m

Try it · work it on paper first

Using the graph below, find the acceleration during C and the total displacement from 0 to 8 s.

Acceleration in C (slope): (−3 − 6) m/s ÷ (8 − 5) s = −3 m/s²

Area of A (triangle): ½ × 2 s × 6 m/s = 6 m. Area of B (rectangle): 3 s × 6 m/s = 18 m.

Area of C: above the axis, ½ × 2 s × 6 m/s = 6 m. Below the axis, ½ × 1 s × 3 m/s = 1.5 m, which counts as −1.5 m.

Add with signs: 6 + 18 + 6 − 1.5 = +28.5 m

Answer: a = −3 m/s² during C; total displacement = 28.5 m in the + direction. (The distance traveled is 31.5 m, because distance counts all the area as positive.)

5. Common mistakes

“Negative means slowing down.”A negative velocity is a direction. An object slows down only when v and a have opposite signs.

Reading the height as position.The height is velocity. A v–t graph alone can’t tell you where the object is.

Adding every area as positive.That gives distance. For displacement, area below the time axis subtracts.

Check yourself

A velocity–time graph is a straight line from −8 m/s up to 0 m/s over 4 s. What is the object doing during those 4 s?

The velocity is negative the whole time, so it moves in the − direction. The graph heads toward the time axis, so its speed drops from 8 m/s to 0. The acceleration is +2 m/s², opposite to the velocity.

Ready to practice?

5 questions on Velocity–Time Graphs, each with a hint and an explanation for every wrong answer.

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