A 30-second animated trip. The car waits, speeds up, cruises, brakes, stops, reverses and stops again, and every change appears on all three graphs at the same moment. Use ⛶ for full screen on a classroom projector.
Pick a preset, set your own starting velocity and acceleration, or drive the car yourself and watch the graphs record. Hover over (or touch) any graph to read values, see the slope (tangent) and the area under the velocity–time line.
Runs for 10 s. Try u = 8, a = −2: the car stops after 4 s, then speeds up backwards.
Hold a button (or use the keyboard: → gas, ← reverse, ↓ or S brake). Let go and the car coasts at constant velocity, as on a frictionless road. You get 20 s.
Challenges: ① draw a “tent” on the position–time graph; ② keep the velocity flat at about 6 m/s for 5 s; ③ finish exactly back at x = 0.
The graphs you read in a test: where the object is at rest, speeding up, moving steadily, slowing down or going back, and what the slope and the area mean. Switch between the labeled diagram, a blank one to test your memory, and a quiz where you place each label yourself. You can also download both versions or print a label worksheet.
Tap a word, then tap the numbered box it belongs to (on the diagram or in the list below). Tap a filled box to take the word back. On a phone, swipe the diagram sideways to see every number.
Ten questions (from a bank of twenty) on slope, area, distance–time and velocity–time graphs.
Imagine filming a car and writing down where it is every second. Plot those positions against time and you get a distance–time graph. If the car is parked, the position never changes, so the line is flat. If it drives steadily, it covers the same distance every second, so the line goes up in a straight slope. Faster car, steeper line. Watch the dots on the ticker tape in the story: far apart when the car is fast, close together when it is slow.
The slope (gradient) of a line is rise ÷ run. On a distance–time graph the rise is a distance and the run is a time, so the slope is speed: 24 m in 4 s is a slope of 6 m/s. A curve that gets steeper means the speed is increasing (accelerating); a curve that flattens means it is slowing down. On a velocity–time graph the same trick gives the acceleration: rising from 0 to 6 m/s in 3 s is 2 m/s². And because distance = speed × time, the area under a velocity–time line is the distance travelled (for motion in one direction).
Velocity has a direction. Use position (displacement) instead of distance and the graph can go down: a falling line means moving back toward the start, with a negative velocity, and on the velocity–time graph the line drops below the axis. Area below the axis is displacement in the negative direction, so the story car goes 42 m forward and 20 m back: displacement 22 m, distance 62 m. In calculus language, velocity is the derivative of position, $v = \frac{dx}{dt}$ (the slope of the tangent), acceleration is $a = \frac{dv}{dt}$, and displacement is the integral $\Delta x = \int v\,dt$ (the area). For constant acceleration these give the equations of motion, $v = u + at$ and $s = ut + \tfrac{1}{2}at^2$.
and watch the slope triangle: rise ÷ run stays 6 m/s. In the lab, draws one straight v–t line that crosses zero while the x–t graph is a hill. Set and hover the velocity–time graph at t = 4 s: the area is 28 m. Then drive it yourself and try to finish exactly at x = 0.
| Graph | Slope means | Area under it means | Flat (horizontal) line means |
|---|---|---|---|
| Distance–time / position–time | speed (velocity) | nothing useful | stationary (at rest) |
| Velocity–time | acceleration | displacement (equal to the distance travelled if the object never turns back) | constant velocity, not stopped |
| Acceleration–time | rate of change of acceleration | change in velocity | constant acceleration |
With starting velocity $u$, final velocity $v$, acceleration $a$, time $t$ and displacement $s$:
Each one is a graph fact: $v = u + at$ is the straight line on a velocity–time graph (intercept $u$, slope $a$), and $s = \tfrac{1}{2}(u + v)\,t$ is the area of the trapezium under it.
On a curved position–time graph the slope changes from moment to moment. The slope of the straight line joining two points (a chord) is the average velocity over that interval, $\bar v = \frac{\Delta x}{\Delta t}$. The slope of the tangent at one point is the instantaneous velocity, $v = \lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}$. Turn on the tangent in the lab and hover along a curve to watch it change.
An object speeds up when its velocity and acceleration have the same sign and slows down when they have opposite signs. In the story, the reversing car (v < 0) with a = −2 m/s² is speeding up, and with a = +2 m/s² it is slowing down. On the v–t graph: moving away from the axis = speeding up, moving toward it = slowing down.
Displacement is the signed area, $\Delta x = \int_{t_1}^{t_2} v\,dt$. Distance travelled is the area of $|v|$: add the areas above and below the axis as positive numbers. The lab shows both when they differ.
Velocity cannot jump instantly: that would need an infinite acceleration. Real position–time graphs are smooth, and the corners in textbook graphs (and in the lab’s “Walk there and back” preset) are idealised. Acceleration can change very quickly (brakes going on), so the story’s a–t steps are a good approximation: the v–t graph may have corners but never gaps.
The slope (gradient) of a distance–time graph is the speed: rise ÷ run = distance ÷ time. A steeper line means a faster object, a horizontal line means the object is stationary, and a curved line means the speed is changing. On a position–time graph the slope is the velocity, so a line sloping down means motion back toward the start.
Key takeaway: slope of distance–time = speed.A horizontal line on a velocity–time graph means the velocity is constant and the acceleration is zero. It does not mean the object is stopped; an object is only at rest when the line lies on the time axis (v = 0).
Key takeaway: flat v–t line = constant velocity, not stopped.Find the area between the line and the time axis. Split the shape into rectangles (base × height) and triangles (½ × base × height) and add them. For example, speeding up from 0 to 20 m/s in 5 s gives ½ × 5 × 20 = 50 m. If part of the line is below the axis, that area is motion in the negative direction: add it as a positive number for the total distance, or subtract it for the displacement.
Key takeaway: distance = area under the velocity–time graph.Acceleration is the slope of the velocity–time graph: a = (change in velocity) ÷ (change in time). A line rising from 0 to 20 m/s over 5 s has a slope of 4 m/s². A downward slope means negative acceleration.
Key takeaway: acceleration = slope of the velocity–time graph.Distance never decreases, so a distance–time graph can only go up or stay flat. Displacement (position) has a direction, so a displacement–time graph slopes down when the object moves back toward its starting point. If an object goes 42 m forward and 20 m back, its distance is 62 m but its displacement is 22 m.
Key takeaway: distance only goes up; displacement can go down.A curved line means the speed is changing. If the curve gets steeper, the object is speeding up (accelerating); if it flattens out, the object is slowing down (decelerating).
Key takeaway: curve = changing speed.Negative velocity (the line below the time axis) means the object is moving in the negative direction, for example back toward its start. It can still be speeding up or slowing down: if the line moves away from the axis it is speeding up, and if it moves toward the axis it is slowing down.
Key takeaway: below the axis = moving the other way.It shows how the acceleration changes with time. For motion made of steady pushes and brakes it is a set of flat steps: above zero while speeding up forward (or braking while reversing), zero at constant velocity, below zero while braking forward (or speeding up in reverse). The area under it is the change in velocity: 2 m/s² for 3 s adds 6 m/s.
Key takeaway: area under a–t = change in velocity.Split the story into parts and draw one piece for each. Standing still: a flat line. Steady speed: a straight sloping line whose slope is the speed. Speeding up: a curve that gets steeper. Slowing down: a curve that flattens out. For a trip back to the start, draw the position line sloping down. Check the total time and distance at the end of each part.
Key takeaway: one piece of graph for each part of the story.Draw the tangent to the curve at that moment and find its slope (rise ÷ run of the tangent line). That slope is the instantaneous velocity, v = dx/dt. The slope of a chord between two points only gives the average velocity over that interval.
Key takeaway: instantaneous velocity = slope of the tangent.