Virginia Research Institute
Virtual Laboratory · Built by E2 Innovations
Explore how position, velocity, and acceleration are linked through real-time graphs. Set initial conditions, run scenarios, drag the car freely, and challenge yourself to reproduce target motion profiles.
How position, velocity, and acceleration relate through graphs and equations.
Position (x) — where the object is relative to the origin. Positive = right of origin, Negative = left. Units: metres (m).
Velocity (v) — rate of change of position. Positive = moving right, Negative = moving left. Units: m/s.
Acceleration (a) — rate of change of velocity. Can be positive or negative regardless of direction of motion. Units: m/s².
Position vs Time (x−t): Slope = velocity. Steep = fast, Flat = stationary, Curve = accelerating. Positive slope = moving right.
Velocity vs Time (v−t): Slope = acceleration. Flat line = constant velocity. Area under the graph = displacement.
Acceleration vs Time (a−t): Flat horizontal line = constant acceleration. Zero line = no acceleration.
For constant acceleration only:
If you know three of the five variables (x, x₀, v, v₀, a, t) you can find the other two using these equations.
Constant velocity (a = 0):
x−t: straight diagonal · v−t: flat · a−t: zero
Constant acceleration:
x−t: parabola · v−t: straight slope · a−t: flat
At rest (v = 0, a = 0):
x−t: horizontal · v−t: zero · a−t: zero
Reversing direction:
v−t crosses zero · x−t reaches peak/trough
Four experiments building from basic to advanced. Complete each one and record your data.
| Time t (s) | Position x (m) | Velocity v (m/s) | Acceleration a (m/s²) |
|---|---|---|---|
| 0 | |||
| 2 | |||
| 4 | |||
| 6 | |||
| 8 |
| t (s) | x (m) | v (m/s) | Δv (m/s) | x predicted = −8 + ½(1.5)t² |
|---|---|---|---|---|
| 0 | — | −8.00 | ||
| 1 | −7.25 | |||
| 2 | −5.00 | |||
| 3 | −1.25 | |||
| 4 | +4.00 |
| Challenge | Description | Your x₀ | Your v₀ | Your a |
|---|---|---|---|---|
| 1 | Constant velocity rightward | |||
| 2 | Accelerate from rest | |||
| 3 | Decelerate to stop | |||
| 4 | Return to starting position |
To investigate the relationships between position, velocity, and acceleration for one-dimensional motion, and to verify the kinematic equations for both constant velocity and constant acceleration using a live simulation. The lab confirms that the slope of a position–time graph equals velocity and that the slope of a velocity–time graph equals acceleration.
For motion along a straight line with constant acceleration, position and velocity are related to the initial conditions by the kinematic equations:
v = v₀ + at
x = x₀ + v₀t + ½at²
On a position–time (x−t) graph the instantaneous slope is the velocity, and on a velocity–time (v−t) graph the slope is the acceleration. When a = 0 the x−t graph is a straight line. When a is constant and non-zero the x−t graph is a parabola and the v−t graph is a straight line. The acceleration–time (a−t) graph is a flat horizontal line at the value a in both cases.
Position equation: x = x₀ + v₀t = −8 + 2t
Position at t = 4 s: x = −8 + 2(4) = 0 m
Velocity: v = v₀ + at = 2 + 0 = +2.00 m/s (constant)
Slope of x−t graph: Δx / Δt = (+8 − (−8)) / (8 − 0) = 16 / 8 = +2.00 m/s ✓ (matches velocity)
Acceleration: a = 0 (v−t graph is flat)
Position equation: x = x₀ + v₀t + ½at² = −8 + ½(1.5)t² = −8 + 0.75t²
Position at t = 4 s: x = −8 + 0.75(16) = −8 + 12 = +4.00 m
Velocity at t = 4 s: v = v₀ + at = 0 + (1.5)(4) = +6.00 m/s
Slope of v−t graph: Δv / Δt = (6 − 0) / (4 − 0) = +1.50 m/s² ✓ (matches acceleration)
Δv per second: constant at +1.50 m/s, confirming constant acceleration.
Experiment 1 — Constant Velocity (x₀ = −8 m, v₀ = +2 m/s, a = 0)
| t (s) | x sim (m) | x = −8 + 2t (m) | v (m/s) | a (m/s²) |
|---|---|---|---|---|
| 0 | −8.00 | −8.00 | +2.00 | 0.00 |
| 2 | −4.00 | −4.00 | +2.00 | 0.00 |
| 4 | 0.00 | 0.00 | +2.00 | 0.00 |
| 6 | +4.00 | +4.00 | +2.00 | 0.00 |
| 8 | +8.00 | +8.00 | +2.00 | 0.00 |
Experiment 2 — Constant Acceleration (x₀ = −8 m, v₀ = 0, a = +1.5 m/s²)
| t (s) | x sim (m) | x = −8 + ½(1.5)t² (m) | v = 1.5t (m/s) | Δv/s (m/s) |
|---|---|---|---|---|
| 0 | −8.00 | −8.00 | 0.00 | — |
| 1 | −7.25 | −7.25 | 1.50 | +1.50 |
| 2 | −5.00 | −5.00 | 3.00 | +1.50 |
| 3 | −1.25 | −1.25 | 4.50 | +1.50 |
| 4 | +4.00 | +4.00 | 6.00 | +1.50 |
The student-calculated positions and velocities agreed with the simulation values exactly for every sampled time in both experiments. In Experiment 1 the x−t graph was a straight line with slope +2.00 m/s, the v−t graph was a flat horizontal line at +2.00 m/s, and the a−t graph was a flat line at zero. This matched the kinematic equations for constant velocity perfectly.
In Experiment 2 the x−t graph curved as a parabola opening upward, the v−t graph was a straight diagonal line with slope +1.50 m/s², and the a−t graph was flat at +1.50 m/s². The change in velocity Δv was constant at +1.50 m/s per second — the defining feature of constant acceleration. The slope of the v−t graph equals the acceleration value set on the control slider, confirming that the first derivative of velocity with respect to time is acceleration.
The most important result across both experiments is that the shape of each graph directly reveals the value of the quantity one level below it: the slope of x−t is v, and the slope of v−t is a. This is the core of graphical kinematics and the reason motion graphs are such a powerful analysis tool.
The experiment successfully verified the kinematic equations for constant velocity and constant acceleration motion. Simulation values matched hand-calculated values to the full precision of the readout in both experiments. The slope of the x−t graph was shown to equal velocity, and the slope of the v−t graph was shown to equal acceleration. Motion graphs are therefore confirmed as an accurate and efficient way to describe, analyze, and predict one-dimensional motion.
Attempt each fully before checking the answer.