Graph Translation & Non-Uniform Acceleration
- Convert between position–time, velocity–time and acceleration–time graphs by slope and area
- Read a curved velocity–time graph, where the kinematic equations no longer apply
- Extract a physical quantity from the slope of a deliberately linearized plot
One motion, three graphs
A single motion can be drawn three ways, and the exam moves freely between them. Going down the chain — x → v → a — you take slopes. Going up the chain — a → v → x — you take areas. Nothing else is needed, and no equation is required, which is exactly why graph questions appear on forms where the algebra would be intractable.
Curvature carries information too
On an x–t graph, a straight line means constant velocity and a curve means the velocity is changing. Curving upward (concave up) means positive acceleration; curving downward means negative acceleration. So you can read the sign of acceleration off a position graph without ever computing a slope — a skill worth having, because it takes two seconds and settles most multiple-choice graph items.
When acceleration is not constant
The three kinematic equations all assume constant acceleration. If a velocity–time graph is curved, the acceleration is changing and those equations are simply wrong — no amount of careful algebra rescues them. But the graph still works: the area under the curve is still the displacement, and the slope of the tangent at a point is still the instantaneous acceleration. Estimating that area by counting grid squares is a legitimate AP method and is often exactly what the rubric asks for.
A cart's velocity–time graph rises linearly from 0 to 8 m/s over the first 4 s, holds at 8 m/s for 3 s, then falls linearly to 0 over the final 2 s. Find the total displacement and the acceleration during each phase.
- 1.Phase 1 is a triangle: area = ½ × 4 s × 8 m/s = 16 m. Slope = 8 ÷ 4 = +2 m/s².
- 2.Phase 2 is a rectangle: area = 3 s × 8 m/s = 24 m. Slope is zero, so a = 0.
- 3.Phase 3 is a triangle: area = ½ × 2 s × 8 m/s = 8 m. Slope = (0 − 8) ÷ 2 = −4 m/s².
- 4.Total displacement is the sum of the three areas: 16 + 24 + 8.
Linearizing: why the exam asks you to plot something strange
Free-response questions routinely ask you to plot data so the result is a straight line, then read a physical quantity from the slope. The move is always the same: rearrange the relationship into y = mx + b form and see what has to go on each axis. For a ball dropped from rest, Δy = ½gt² is a parabola against t — but plot Δy against t² and it becomes a straight line of slope ½g. The slope then hands you g, and a best-fit line through many data points is far more reliable than any single measurement.
A high point on a velocity–time graph does not mean the object is far away, and a v–t graph crossing zero does not mean the object is at the origin. Height on a v–t graph is speed; the area is the only thing that tells you about position.
A position–time graph is a curve that is concave down and has positive slope everywhere shown. What is happening?
An object starts at rest and its acceleration–time graph is a constant +3 m/s² for 4 s. What is its velocity at t = 4 s?
A student drops a ball from several heights and plots fall height Δy against the square of the fall time, t². The best-fit line has slope 4.9 m/s². What quantity has she measured?
If a free-response part says "the students plot the data so that the graph is linear", it is asking you to rearrange the equation into y = mx + b and name the axes. State what goes on each axis and what the slope represents — both are separate rubric points.
Answer the 3 checkpoints as you read.
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