AP PHYSICS 1 AND 2 › UNIT 1, KINEMATICS › TOPIC 1.3
Representing Motion
One motion, four ways of showing it: a diagram of dots, a position graph, a velocity graph and an acceleration graph. They are not four topics. They are one motion, and this topic is about moving between them without losing your footing.
1.3.A Represent the motion of an object with a diagram and with graphs of position, velocity and acceleration against time.
1.3.B Move between those representations, and say what a slope or an area means on each one.
1. The motion diagram, before any axes
Take a photograph every second and mark where the object was. That row of dots is a motion diagram, and it carries more than it looks like it does. Evenly spaced dots mean a steady speed. Dots that spread out mean speeding up. Dots that bunch together mean slowing down.
Nothing on a motion diagram is a number. That is the point of drawing one first: it forces you to decide what the motion is before you decide what it measures.
2. Slope on a position graph is velocity
On a graph of position against time, the slope at an instant is the velocity at that instant. Steeper means faster. Sloping down means moving in the negative direction. Flat means stopped.
3. On a velocity graph, the slope is acceleration and the area is displacement
The velocity graph carries two different pieces of information, read in two different ways, and keeping them apart is most of the skill in this topic.
slope of v against t = accelerationarea under v against t = displacement
The area is signed. Area below the axis counts as negative, which is what makes the total a displacement rather than a distance.
Build a motion and watch all four representations at once
Drag the round handles on the velocity graph to invent a motion. Everything else redraws as you drag: the acceleration below it is the slope of what you are holding, the position above it is the running area underneath, and the dots at the top are the motion diagram. Drag the time cursor to shade the area up to that moment.
Two things are worth hunting for by hand in the starting motion. The object is at rest at t = 0 and again at t = 5.5 s, and the second one is where the position graph turns over, because that is where the area stops being added and starts being taken away. And the acceleration graph is a set of flat steps, because a velocity graph made of straight lines has a constant slope on each one.
4. On an acceleration graph, the area is the change in velocity
The same reading applies one level down. The area under an acceleration graph between two times is how much the velocity changed over that interval. Not the velocity. The change.
area under a against t = change in velocity
So an acceleration graph alone can never tell you how fast something is going, in the same way a velocity graph alone can never tell you where something is. Each graph is missing exactly one number, and that number is the starting value.
5. Reading between the graphs, in order
When a question hands you one graph and asks about another, the same four questions work every time, and in this order.
Where is it zero? A zero on the velocity graph is a flat spot on the position graph. A zero on the acceleration graph is a flat spot on the velocity graph.
Where does it change sign? That is where the graph above it turns around.
Is the slope positive or negative? That gives you the sign of the graph below it.
Is the area piling up or cancelling? That gives you the graph above it.
Match the graphs
A velocity graph is shown. Three position graphs are offered and one of them goes with it. Pick one, and the feedback tells you what to look at rather than just whether you were right.
Check yourself
1. A velocity graph is a straight horizontal line at +4 m/s for 6 s. Describe the position graph and the acceleration graph.
Position: a straight line sloping up with a slope of 4, rising by 24 m over the six seconds. Acceleration: a flat line on zero. The acceleration is zero even though the object is moving quickly, which is the whole content of the question.
2. In the starting motion of the simulation, the velocity is +6 m/s from t = 2 to t = 4 s. Find the displacement over that interval two ways, and say where the object is at t = 4 s if it started at the origin.
Area of the rectangle: 6 × 2 = 12 m. Or read the position graph: it goes from 6 m to 18 m, a change of 12 m. At t = 4 s the object is at 18 m. Reading the change off the position graph and computing the area under the velocity graph are the same operation.
3. The same motion has the object momentarily at rest at t = 5.5 s. Say what is happening to the position and to the acceleration at that instant.
The position is at its maximum, 22.5 m, and the position graph is momentarily flat. The acceleration is −4 m/s², not zero: it is what turns the object around. At rest and not accelerating are different statements, and only one of them is true here.
4. Over the whole eight seconds the object ends at +18 m. Is the distance it traveled also 18 m? Work it out.
No. It went forward to 22.5 m, then came back to 18 m, so it covered 22.5 + 4.5 = 27 m. The displacement is the signed area, +18 m. The distance adds the areas as positive numbers regardless of which side of the axis they are on.
Topic 1.4, Reference Frames and Relative Motion. Every graph on this page was drawn from somebody’s point of view. The next topic asks what happens to all of them when the person holding the stopwatch is moving too.