Simulation

Constant Acceleration Cart

KinematicsAcceleration

A cart with constant acceleration; live position, velocity, and acceleration plots update side by side

Objective

Verify that a cart under constant acceleration obeys x = v₀·t + ½·a·t² and v = v₀ + a·t — the two fundamental kinematic equations. The sim treats the cart as a point mass on a frictionless horizontal track with no air resistance, so both equations hold exactly throughout the run.

Setup

Set the Acceleration slider to 2 m/s² and Initial Velocity to 0 m/s. These are the default values; confirm the Position readout reads 0.00 m.
Press Start and observe the cart moving rightward along the track. Watch the Time, Position, and Velocity readouts updating each frame.
At t = 3 s, record the Position and Velocity readouts. The predicted values are x = 9.00 m and v = 6.00 m/s.
Press Reset, then set Acceleration to −2 m/s² and Initial Velocity to 8 m/s. Press Start and observe the cart decelerating — the velocity readout should reach zero near t = 4 s.
Let the simulation run to its natural stop at t = 10 s. Confirm the final position matches the analytical prediction for that parameter set.

Analytical Prediction

With v₀ = 0 m/s and a = 2 m/s², the kinematic equations predict position and velocity at t = 3 s:

x=v₀·t + ½·a·t²

=0 + ½·2·(3)²

=9.00 m

v=v₀ + a·t

=0 + 2·3

=6.00 m/s

For the decelerating case (v₀ = 8 m/s, a = −2 m/s²), the cart reaches rest when v = 0 at t = 4 s:

x_stop=v₀·t + ½·a·t²

=8·4 + ½·(−2)·(4)²

=32 − 16

=16.00 m

After t = 4 s the cart reverses direction and returns leftward, reaching x = 0 again at t = 8 s.

Results Analysis

Compare the Position (m) readout to x = ½·a·t² at t = 3 s — the readout should show 9.00 m within ±0.02 m. Compare the Velocity (m/s) readout to v = a·t — it should show 6.00 m/s within ±0.02 m/s. The right-panel x(t) plot curves upward as a parabola while v(t) rises as a straight line; both shapes confirm the kinematic law visually. The a(t) trace remains a flat horizontal line at the set value, confirming the acceleration is truly constant. For the decelerating run, the v(t) line crosses zero at t ≈ 4 s and x(t) peaks near 16 m before declining.

Source of Error

The simulation models a point mass on a perfectly frictionless, horizontal track with no air resistance, no rolling losses, and no rotational inertia — the same idealizations assumed in the analytical prediction. Real carts experience rolling friction, axle bearing drag, and aerodynamic resistance proportional to velocity, none of which appear here. The analytical prediction uses the identical frictionless, drag-free model, so both sides share the same idealizations and they cancel. The residual gap between the readout values and the prediction is therefore purely numerical, not physical.

Further Exploration

Set acceleration to 0 m/s² and initial velocity to 5 m/s. Does the cart travel at constant speed? What do the v(t) and x(t) plots look like — and why is this the limiting case of the kinematic equation?
What acceleration value makes the cart travel exactly 50 m in 10 s starting from rest? Use x = ½·a·t² to compute it first, then verify by running the sim.
Set a negative acceleration with a large positive initial velocity. At what time does the velocity readout cross zero? Does the position peak match the analytical stopping distance v₀²/(2|a|)?
Try the maximum acceleration of 5 m/s². How does the curvature of the x(t) plot change compared to a = 1 m/s²? The curvature of a parabola x = ½·a·t² is exactly a — can you see that relationship in the plots?