B. Kinematics · Kinematics: Motion in One and Two Dimensions

Moving Between Graphs: Position-Time, Velocity-Time and Acceleration-Time

B2.2 B3.1 A1.6About 70 minutes

Learning goals

I can read an acceleration-time graph and use its area to find the change in velocity.
I can sketch the v–t graph from a d–t graph using slopes.
I can sketch the a–t and d–t graphs from a v–t graph using slopes and areas.
I can describe the matching shapes of the three graphs for rest, constant velocity and constant acceleration.

Words to know

acceleration-time graph (a–t graph)
A graph with time on the horizontal axis and acceleration on the vertical axis.

area under an a–t graph
Acceleration × time = change in velocity (Δv).

parabola
The U-shaped curve of a quadratic relation. The d–t graph of constant acceleration is part of a parabola.

concave up / concave down
A curve that bends upward (getting steeper) / bends downward (getting less steep).

Three views of one trip

A weather app can show the same day as a temperature graph, a rain graph or a wind graph. Each tells you something the others hide. Motion works the same way. One trip can be drawn as a position-time, a velocity-time or an acceleration-time graph. In this lesson you will learn to translate between them, which is one of the most useful skills in the course.

The two links: slope and area

Going down the list (d → v → a), use slopes:

slope of d–t = velocity
slope of v–t = acceleration

Going up the list (a → v → d), use areas:

area under a–t = change in velocity, Δv
area under v–t = change in position, Δd

An acceleration-time (a–t) graph is usually simple in this course. For uniformly accelerated motion the acceleration is constant, so the a–t graph is a horizontal line (it may jump to a new value when the motion changes).

The matching shapes for three kinds of motion in the positive direction. Each graph is the slope of the one to its left. For constant acceleration, the d–t graph is a curve (part of a parabola) that gets steeper.

Worked example

Example 1: From a v–t graph to the a–t and d–t graphs

A cart starts from rest at d = 0. Its v–t graph is a straight line from (0 s, 0) to (2 s, 6 m/s), horizontal to (5 s, 6 m/s), then a straight line down to (8 s, 0), and it stays at rest until 9 s. Sketch the a–t and d–t graphs, giving key values.

a–t (slopes of v–t): 0–2 s: (6 − 0) ÷ 2 = 3 m/s². 2–5 s: horizontal, 0. 5–8 s: (0 − 6) ÷ 3 = −2 m/s². 8–9 s: 0. Draw a horizontal segment at each value.
Positions (areas of v–t): 0–2 s: ½(2)(6) = 6 m, so d(2) = 6 m. 2–5 s: (3)(6) = 18 m, so d(5) = 24 m. 5–8 s: ½(3)(6) = 9 m, so d(8) = 33 m. After 8 s it stays at 33 m.
d–t shapes: 0–2 s the velocity grows, so the d–t graph curves and gets steeper (concave up). 2–5 s the velocity is constant: a straight line of slope 6. 5–8 s the velocity drops to 0: the curve gets flatter (concave down) until it is horizontal at 8 s. 8–9 s: horizontal.
Check: the slope of the d–t graph at each time should equal the v–t value. At 8 s the d–t graph is flat, matching v = 0. ✓

a–t: 3, 0, −2, 0 m/s². d–t passes through (2 s, 6 m), (5 s, 24 m) and (8 s, 33 m), curved–straight–curved–flat.

Example 1 drawn three ways on the same time axis. The dashed lines mark where the motion changes. Notice how the d–t curve bends upward while the cart speeds up and bends over while it slows down.

Worked example

Example 2: Area under an a–t graph

A cyclist moving at 3.0 m/s [N] accelerates at a constant 2.5 m/s² [N] for 4.0 s. Use an a–t graph to find her final velocity.

The a–t graph is a horizontal line at 2.5 m/s² from 0 to 4.0 s. The area is a rectangle.
Δv = area = 2.5 m/s² × 4.0 s = 10 m/s [N].
v₂ = v₁ + Δv = 3.0 + 10 = 13 m/s [N].
Note: the a–t graph gives only the change in velocity. You must know v₁ to find v₂ (just as a v–t graph gives Δd, and you must know the starting position to find the final position).

13 m/s [N]

Worked example

Example 3: From a d–t graph to a v–t graph

A d–t graph (east positive) has straight segments from (0 s, 0 m) to (3 s, 12 m), then to (5 s, 12 m), then to (7 s, 0 m). Sketch the v–t graph.

Find each slope. 0–3 s: 12 ÷ 3 = 4 m/s. 3–5 s: 0. 5–7 s: (0 − 12) ÷ 2 = −6 m/s.
Each straight d–t segment becomes a horizontal v–t segment: at 4 m/s from 0 to 3 s, at 0 from 3 to 5 s, and at −6 m/s from 5 to 7 s.
Check with areas: 4 × 3 = 12 m, then 0, then −6 × 2 = −12 m. Total Δd = 0, and the object ends where it started, as the d–t graph shows. ✓
(Real objects cannot change velocity instantly, so the jumps on a real v–t graph would be very steep lines rather than vertical ones. The idealized graph is fine for problem solving.)

v–t: 4 m/s (0–3 s), 0 (3–5 s), −6 m/s (5–7 s).

A checklist for sketching

Read the axis labels first. Is this d–t, v–t or a–t?
Split the time axis wherever the motion changes. Line up all your graphs on the same time scale.
Going down a level, ask "what is the slope here?" Going up a level, ask "what is the area so far?"
Mark key values with numbers (turning points, where graphs cross zero).
Check: zero velocity on the v–t graph must line up with a flat spot on the d–t graph.

Just for fun

A position-time graph, a velocity-time graph and an acceleration-time graph walk into a café.

The server looks at them and says, "Let me guess. You're all here to see how things are trending."

Hands-on activity

Walk the graph

You need: the floor number line from Lesson 1 (or a hallway with metre marks), a stopwatch, a partner and paper.

Your partner draws a simple v–t graph with three parts, for example: 0.5 m/s for 4 s, stop for 2 s, then −1.0 m/s for 2 s.
Before walking, sketch the matching d–t graph and work out where you should end up (here: +2.0 m, then +2.0 m, then 0 m).
Walk it while your partner calls out the seconds. Did you finish at the predicted position?
Swap roles. Then try a v–t graph with a gentle speed-up (a sloped line) and describe how your walking felt.

What to look for: areas on the v–t graph predict the end position. For the sample graph, 0.5 × 4 = 2.0 m, then 0, then −1.0 × 2 = −2.0 m, so you end back at 0. A sloped v–t segment should feel like a steady speed-up, and the d–t sketch should curve there.

Your turn

What is the area under an a–t graph equal to?

A v–t graph is a horizontal line at −3 m/s. Describe the a–t graph and the d–t graph.

A v–t graph is a straight line from (0, 0) to (4 s, 8 m/s). Sketch the a–t graph.

For the same motion, starting at d = 0, what is the position at 4 s, and what shape is the d–t graph?

A d–t graph is a straight line from (0, 20 m) to (5 s, 0). Sketch the v–t graph.

An object's a–t graph shows 1.5 m/s² for 6.0 s. It started at 2.0 m/s. Find its final velocity.

More practice: warm-up

Which graph's slope gives acceleration?

For uniform motion, what does the a–t graph look like?

For constant acceleration, what does the v–t graph look like?

For constant acceleration from rest, what does the d–t graph look like?

A flat (horizontal) d–t graph matches what value on the v–t graph?

More practice: on your own

A v–t graph goes in straight segments through (0, 10 m/s), (4 s, 10 m/s) and (9 s, 0). Sketch the a–t graph with values.

For the same motion, starting at d = 0, find d at 4 s and 9 s and describe the d–t shape.

An a–t graph shows −3.0 m/s² from 0 to 2.0 s, then 0 from 2.0 s to 5.0 s. The object starts at 9.0 m/s. Find its velocity at 2.0 s and 5.0 s.

A d–t graph is a curve that gets steeper and steeper. What do the v–t and a–t graphs show (assume constant acceleration)?

Explain why you cannot find an object's position from a v–t graph alone.

Challenge

A ball is thrown straight up at 9.8 m/s and caught at the same height 2.0 s later (up positive, a = −9.8 m/s²). Sketch all three graphs from 0 to 2.0 s with key values.

A v–t graph goes in straight lines through (0, 0), (3 s, 6 m/s), (6 s, −6 m/s). Find the a–t values and the position at 6 s if d(0) = 0.

Can a d–t graph have a sharp corner (like a V) in real life? What would that mean on the v–t and a–t graphs?

Check your understanding

Quiz

The slope of a d–t graph gives the:

The area under an a–t graph gives the:

For an object with constant positive acceleration, the a–t graph is:

A v–t graph is a horizontal line at 5 m/s from 0 to 6 s. The matching d–t graph (starting at 0) is:

A d–t graph curves and becomes less steep until it is horizontal. On the v–t graph this looks like:

A v–t graph is a straight line from (0, 12 m/s) to (4 s, 0). The a–t graph is:

For parents and teachers

Big idea: The three motion graphs are linked by slopes (going d → v → a) and areas (going a → v → d). This is expectation B2.2 in full and is excellent preparation for calculus (MCV4U).

How to help: Cover two of the three stacked graphs in Example 1 and ask your teen to rebuild them from the third. Ask "slope or area?" each time.

Common difficulties:

Copying the shape of one graph onto the next (for example drawing a sloped line on the a–t graph because the v–t graph slopes).
Forgetting that areas give changes, so a starting value is needed.
Not lining graphs up on the same time axis.
Drawing the d–t graph with sharp corners where the velocity changes smoothly.