Derive the equations and test their assumptions

1. Orient: Derive the equations and test their assumptions

Constant acceleration means equal time intervals produce equal velocity changes. Starting from a=(v−v₀)/t gives v=v₀+at. The velocity graph is then a straight line, so its average height is (v₀+v)/2. Multiplying that height by time gives displacement Δx=(v₀+v)t/2.

2. Explain the model

Substitute v=v₀+at to obtain Δx=v₀t+0.5at². Alternatively multiply v−v₀=at by v+v₀ and use (v+v₀)t=2Δx to get v²−v₀²=2aΔx. Each relationship uses the same constant-acceleration assumption. The squared-velocity equation loses the sign of final velocity, so a sign choice still needs a motion argument.

The required framework asks for qualitative reasoning about nonuniform acceleration; the following supplied piecewise-history arithmetic is optional enrichment. Constant-acceleration equations and geometric graph interpretation remain core. An average acceleration is not a complete history. A cart can gain all its speed early and coast, gain speed uniformly, or wait before gaining speed late. Equal initial and final velocities give equal average acceleration, but the velocity–time areas differ. The experiment compares these histories without asking for calculus or a force mechanism.

At a turning point v=0 does not imply a=0. If constant acceleration continues, velocity changes sign. A physical brake that switches off or a cart that collides at the turning point would create a different model. Always state whether the rule continues after an event rather than silently stopping the calculation.

3. Work an example

Three carts start at rest, reach 6 m/s after 3 s, and have average acceleration 2 m/s². Uniform acceleration gives displacement 9 m. Accelerating at 4 m/s² for the first 1.5 s then coasting gives 4.5+9=13.5 m. Waiting 1.5 s then accelerating gives only 4.5 m. Shared endpoints do not determine area.

Acceleration-area and qualitative graph translation

For a constant acceleration of −2 m/s² over 3 s, the signed acceleration–time rectangle is −6 m/s, so an initial velocity +4 m/s becomes −2 m/s. Doubling duration doubles the velocity change, not necessarily the distance. For a qualitatively increasing positive acceleration, sketch an increasing velocity graph whose slope becomes steeper and a position graph with increasingly positive curvature. Do not apply one constant-acceleration displacement equation to the whole nonuniform interval.

4. Predict and investigate

Predict first: Which of the three histories travels farthest, and why can their final velocities still be identical?

Choose total velocity change 8 m/s and elapsed time 4 s. Derive all three displacements from geometric areas before using the output. Then halve elapsed time while keeping the velocity change fixed. Distinguish the change in average acceleration from the change in displacement. Use the lesson experiment controls and retain the settings with your evidence.

5. Transfer and explain

Try a new practice case

A cart has initial velocity −2 m/s and constant acceleration +3 m/s² for 4 s. Find its signed displacement.

Your numerical result (m) Check number

Reveal worked check (practice, not mastery)

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Reflect and retain evidence

A motion record provides only initial and final velocity and elapsed time. Which additional evidence would justify the constant-acceleration displacement equation?

Record a claim, numerical or graphical evidence, assumptions, a limitation and a changed-case prediction. A supplied synthetic investigation is not evidence that you performed a physical laboratory.

Framework and source notes

Original teaching prose, cases and synthetic datasets. Framework mapping: 1.3.A. College Board AP Physics 1 CED, Unit 1, checked 30 October 2026. This mapping is a content cross-reference, not AP authorization or a claim of full course completion.