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Kinematics Graphs

Interpret position-time and velocity-time graphs for uniform acceleration

A ball sits on a perfectly level track, and nothing about its ride will ever involve a hill. Yet the moment you push it, three graphs start drawing mountains. The green position curve climbs, peaks, and falls; the blue velocity line tilts toward zero, crosses it, and dives negative; the orange acceleration line never moves at all. Three windows onto one motion — and the oldest trap in physics class is reading the first window as a photograph of the trip. It is not a picture. The vertical axis is position along the track, and a peak on that axis means the ball ran out of forward motion at its farthest point, paused for one instant with velocity zero while the pull kept pulling, and came back the way it came — all on flat ground. In this lab you set the push and the pull yourself, watch all three graphs argue it out in real time, and freeze the single frame where the story turns: the zero crossing of the v-t line, the apex of the x-t curve, and a ball changing its mind on a level rail. By the end, a graph will read to you as a sentence about direction and slope — never as a snapshot of scenery.

What you'll be able to do

  • Read a position-time graph as position-versus-time rather than as a picture of the path: identify turnarounds at extrema, constant velocity in straight sections, and direction of motion in the sign of the slope — on graphs of one-dimensional motion along a level track
  • Translate between the three graph families: the slope of x-t gives instantaneous velocity, the slope of v-t gives acceleration, and the zeros of the v-t graph locate the extrema of the x-t graph — verifying the chain numerically against the sim's readouts (Braking preset: v = 8 − 2t crosses zero at t = 4.0 s exactly where x = −8 + 8t − t² peaks at +8 m)
  • Compute position, velocity, and kinetic energy from the constant-acceleration equations (v = v₀ + at, x = x₀ + v₀t + ½at², KE = ½mv² with the sim's hard-coded m = 1.0 kg) and match each closed-form value to the panel's live readouts

Formulas

x=x0+v0t+12at2x = x_0 + v_0 t + \frac{1}{2}at^2
Position as a function of time (uniform acceleration)
v=v0+atv = v_0 + at
Velocity as a function of time
v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x
Velocity-displacement kinematic equation
Δx=area under v-t graph\Delta x = \text{area under } v\text{-}t \text{ graph}
Displacement equals the area under the velocity-time graph

Make a prediction

Load the Braking preset: the ball starts at x = −8 m on a straight, level track, launched at +8 m/s with a constant pull of −2 m/s². The green x-t graph will climb to a peak and curve back down. Before you run it — at the peak of that x-t curve (t = 4.0 s), where is the ball and what is it doing?

No grading here — commit to a guess, then scroll down and test it yourself.

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Check

Did the lab agree with you?

Your prediction

You skipped the prediction — jump back to the preview and commit to a guess first; the comparison is the whole point.

Answer: It is momentarily stopped at x = +8 m — the farthest point of its trip along the level track — with acceleration still at −2 m/s², about to roll back toward −x. The peak of the x-t graph is a turnaround in position, not a hilltop

Position is x(t) = −8 + 8t − t² and velocity is v(t) = 8 − 2t, so at t = 4.0 s: v = 0 and x = −8 + 32 − 16 = +8 m — the ball's maximum position, paused for one instant before the still-active −2 m/s² pull sends it back toward −x. The sim confirms it three ways at once: dV flips through 0.00 m/s, dA never leaves −2.00 m/s², and the ball visibly reverses at the x-t apex. Option B is the targeted misconception — the graph-as-picture error, the most documented misreading of kinematics graphs (a research-based distractor family in Beichner's TUG-K, named verbatim in Klein et al. 2019, traced to McDermott, Rosenquist & van Zee 1987). The decisive witness is the sim itself: the track is level and the ball's height is hard-coded constant (y = 1.1 every frame) — the curve rises while the ball never climbs, because the vertical axis is position along the track, not altitude. Option C smuggles in a second documented error: zero velocity does not mean zero acceleration — the pull does not pause when the motion does (the sim's own misconception card states it for a thrown ball: v = 0 at the top, a = −9.8 m/s² throughout).

Quiz (0/3)

A car starts at 4 m/s and decelerates at -2 m/s². When does it come to a stop?

What does a negative area under a v-t graph represent physically?

Sketch what an a-t graph looks like for constant acceleration motion. What shape is it?

You can now

  • Read a position-time graph as position-versus-time rather than as a picture of the path: identify turnarounds at extrema, constant velocity in straight sections, and direction of motion in the sign of the slope — on graphs of one-dimensional motion along a level track
  • Translate between the three graph families: the slope of x-t gives instantaneous velocity, the slope of v-t gives acceleration, and the zeros of the v-t graph locate the extrema of the x-t graph — verifying the chain numerically against the sim's readouts (Braking preset: v = 8 − 2t crosses zero at t = 4.0 s exactly where x = −8 + 8t − t² peaks at +8 m)
  • Compute position, velocity, and kinetic energy from the constant-acceleration equations (v = v₀ + at, x = x₀ + v₀t + ½at², KE = ½mv² with the sim's hard-coded m = 1.0 kg) and match each closed-form value to the panel's live readouts

What you'll learn

  • Position-Time Graphs. A position-time graph shows where an object is at each moment. For constant acceleration, the curve is a parabola. The slope at any point equals the instantaneous velocity — steeper slope means faster motion.
  • Velocity-Time Graphs. A velocity-time graph for constant acceleration is always a straight line. The slope of this line equals the acceleration, and the area under the line equals displacement — not distance traveled.
  • Acceleration-Time Graphs. For uniform acceleration, the a-t graph is a horizontal line. The area under the a-t curve over a time interval equals the change in velocity during that interval. Zero area means the speed did not change.
  • Connecting the Three Graphs. The three motion graphs are mathematically linked: the slope of x-t gives v-t, and the slope of v-t gives a-t. Working backwards, the area under a-t gives change in v, and the area under v-t gives displacement. Mastering these connections is the single most tested skill on AP Physics 1.

Step-by-step

  1. Adjust initial velocity and acceleration sliders.
  2. The animation shows a ball moving along a track while both graphs update in real time.
  3. Pause at any moment to read the exact x and v values.
  4. Try v₀ = 0, a = 2 to get pure parabolic x-t and linear v-t.
  5. Then try v₀ = 5, a = -2 — note the x-t maximum occurs exactly where v-t crosses zero.
  6. Use the speed buttons (0.25x–3x) to slow down or speed up the run.

Key formulas

  • x=x0+v0t+12at2x = x_0 + v_0 t + \frac{1}{2}at^2Position as a function of time (uniform acceleration)
  • v=v0+atv = v_0 + atVelocity as a function of time
  • v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta xVelocity-displacement kinematic equation
  • Δx=area under v-t graph\Delta x = \text{area under } v\text{-}t \text{ graph}Displacement equals the area under the velocity-time graph

Frequently asked questions

A car starts at 4 m/s and decelerates at -2 m/s². When does it come to a stop?
The correct answer is: 2 s. You can work it out this way: set v = 0 in v = v₀ + at. Solve for t.
What does a negative area under a v-t graph represent physically?
The correct answer is: The object moved in the negative direction. Area = displacement. Negative area means the object moved in the negative direction.
Sketch what an a-t graph looks like for constant acceleration motion. What shape is it?
The correct answer is: A horizontal line. If acceleration is constant, what does its graph vs. time look like?