AP PHYSICS C: MECHANICS › UNIT 1, KINEMATICS › TOPIC 1.3
Representing Motion
Topic 1.2 gave you the derivative and the integral. This topic is what they look like when nobody hands you a function: three graphs stacked on one time axis, where differentiating means reading a slope and integrating means reading an area.
Move between graphs of position, velocity and acceleration without a formula, using slope in one direction and area in the other.
Read concavity on a position graph and say what it tells you about the acceleration.
Say what each graph cannot tell you on its own, and why the missing piece is an initial condition.
Wording here is mine, written from the Cornell notes. Check it against your own copy of the CED.
1. The two operations, drawn instead of computed
Everything in this topic is one of two readings, and they run in opposite directions.
v = dx/dt, so v is the slope of the x graphΔx = ∫v dt, so Δx is the area under the v graph
The same pair connects velocity to acceleration one level down. Differentiating moves you down the stack and is a question about steepness. Integrating moves you up the stack and is a question about accumulated area. Neither one needs a formula.
2. Concavity is the acceleration, visible on the position graph
On a position graph the slope is the velocity, and the way the slope is changing is the acceleration. That is concavity, and it is worth naming because it lets you read a straight off the top graph without going through the middle one.
Concave up, holding water, means the slope is increasing, so a > 0. Concave down means the slope is decreasing, so the acceleration is negative. A straight line has no curvature at all, so the acceleration is zero however steep it is.
Build a motion and watch all four representations at once
Drag the round handles on the velocity graph. The acceleration below is its derivative, taken segment by segment, and the position above is its integral, accumulated from the left. The dots along the top are where the object actually is at each whole second. Drag the time cursor to shade the area up to that instant.
In the starting motion the velocity is piecewise linear, so the acceleration is a step function and the position is piecewise quadratic. Watch where the pieces meet: the acceleration jumps there, the velocity has a corner, and the position stays perfectly smooth. Each level up integrates away one degree of roughness, which is the reason position graphs of real motions look so much tamer than the forces behind them.
3. What each graph refuses to tell you
Integrating recovers a function only up to a constant, and on a graph that constant is a vertical shift. So a velocity graph fixes every change in position and says nothing about where the motion started.
x(t) = x0 + ∫0t v dtv(t) = v0 + ∫0t a dt
This is why a graph question that asks “where is the object at four seconds” is unanswerable from a velocity graph alone, while “how far did it move between one and four seconds” is answerable immediately. One asks for a value and the other asks for a change.
4. Reading between the graphs, in order
Zeros of the lower graph are stationary points of the upper one. Where v = 0 the position graph is flat; where a = 0 the velocity graph is flat.
Sign changes of the lower graph are turning points of the upper one. A velocity that crosses zero is a position that turns around. An acceleration that crosses zero is a velocity at a maximum or minimum, which is an inflection point on the position graph.
Signed area accumulates upward. Area above the axis adds, area below subtracts, and the running total is the graph one level up.
Match the graphs
A velocity graph is shown and three position graphs are offered. One of them is its integral. The wrong two are the mistakes that actually get made, so read the feedback even when you are right.
Check yourself
1. A position graph is concave down everywhere and its slope is negative throughout. Describe the velocity and the acceleration, and say whether the object is speeding up.
Negative slope means the velocity is negative throughout. Concave down means the acceleration is negative throughout. Same sign, so the object is speeding up the whole time while moving in the negative direction. Anyone who answers “slowing down, because the acceleration is negative” has read the sign and skipped the product.
2. In the starting motion of the simulation, use the graph alone to find the displacement between t = 0 and t = 2 s, and then between t = 6 and t = 8 s.
From 0 to 2 the velocity rises linearly from 0 to 6, so the area is a triangle: ½ × 2 × 6 = +6 m. From 6 to 8 the velocity is a constant −2 m/s, so the area is −2 × 2 = −4 m, below the axis and therefore negative. Both are readable without a formula.
3. Sketch the acceleration graph for a position graph that is a smooth parabola opening downward. Then say where the velocity is zero and what the acceleration is doing there.
A downward parabola has constant negative curvature, so the acceleration is a horizontal line at a negative value: constant, not zero, and not changing. The velocity is zero at the vertex, which is the single instant the object is at rest, and the acceleration is unchanged there. This is the free fall graph, and the vertex is the top of the flight.
4. Two objects have identical velocity graphs. What can you say about their position graphs, and what can you not say?
The position graphs are identical in shape: same slope everywhere, same concavity everywhere, same displacement over every interval. What you cannot say is where either of them is, because the graphs may be separated by any vertical shift at all. That shift is the constant of integration, and only an initial condition fixes it.
Topic 1.4, Reference Frames and Relative Motion. Every graph here was drawn by an observer who was standing still. The next topic asks what all of them look like to somebody who is not.