Physic Labs

Newtonian mechanics

Uniformly accelerated motion

Study straight-line motion at constant acceleration. Verify v=v0+atv = v_0 + at and x=x0+v0t+12at2x = x_0 + v_0t + \frac{1}{2}at^2 via the linear v–t graph whose slope equals the acceleration.

Middle school

Equipment

  • 3D model of a body moving along a straight line with a trail
  • Time slider to sweep through the motion
  • Parameter slider adjusting the acceleration
  • v–t and x–t graphs with a readout line

Procedure

  1. Read the slope of the v–t graph

    Run the simulation and drag the time slider: on the v–t graph, velocity rises along a straight line. Measure the slope from any two points and compare it with the acceleration readout — the slope equals aa.

  2. Check the distance under the graph

    At an instant t chosen with the slider, read the position x on the canvas. The trapezoid area under the v–t line up to that instant equals v0t+12at2v_0t + \frac{1}{2}at^2 — precisely the displacement x−x0x - x_0; compare the two computations.

  3. Change acceleration and watch the curvature

    Drag the parameter slider to raise a: the v–t line steepens and the x–t curve bends harder. Try negative a and watch the body slow then reverse; predict the stopping instant from v=v0+at=0v = v_0 + at = 0 before checking on the figure.

Simulation

Experiment history

In 1604, Galileo Galilei recognized from inclined-plane experiments that distance fallen grows as the square of time — the basis of uniformly accelerated motion. He published the "odd-number" theorem (distances in equal intervals are in ratio 1:3:5:7…) in the Discorsi (1638). Nicole Oresme in the fourteenth century had already sketched velocity–time graph ideas (the geometry of motion), but only with Galileo came experimental measurement. Newton later placed acceleration inside his laws of dynamics (1687), making v=v0+atv = v_0 + at a consequence of constant force.

Related physicists

Galileo GalileiIsaac Newton

Related library topics

Uniformly accelerated motion