A. Introduction

  1. A 1.0-kg cart on a track experiences several different constant accelerations as shown in the animations (position is given in meters and time is given in seconds).  The red dot shows you where position measurements can be taken if desired. 
  2. Below are the three most-used constant acceleration formulas:

    v = v0 + at, 

    x = x0 + v0t + 0.5at2,


    v2 = v02 + 2a(x - x0).

    Before we begin, play each animation with the velocity calculation checked,


B. Animations

  1. Animations 1 and 2 are the case of zero acceleration. It implies that the velocity remains constant as can be seen in these two animations.
  2. Animations 3 and 4 correspond to the case of non zero but constant acceleration. Here the velocity increases as a function of time in a linear fashion according to the equations of motion of the introduction. 
  3. Animation 5 is the case of a constant deceleration. This has the effect of decreasing the magnitude of this object's velocity in a linear manner.

C. Simulations


The first simulation, deals with a particle in a box with walls of adjustable hardness. The idea here is to think about the behavior of the the particle as it enters the wall regions. The particle starts with with an initial velocity. As it enters the wall regions, the model makes use of a decelerating force, not unlike that of a rocket fired in order to accelerate or decelerate. Here, the decelerating force brings the particle to a stop, it makes turn around, and finally accelerates it, bringing back to its original but opposite velocity, thus concerving energy. This model is made possible by solving Newton's 2nd law in differential form: 

m*d2x(x)/dt2 =F(x) 

with boundary conditions such that,

F(x)= {-a1 if x >= L, or a1 if x <=0},

where a1 is a constant and L is the size of the box. The model conserves energy as can be seen in the simulation1, since its kinetic energy, 

E=(1/2)*m*v 2 

remains constant. This is easily seen also by noticing that the particle's speed inside the box is constant as well. The equation is solved by a Runge-Kutta solver using EJS.


The second simulation is one that deals with the interation of two atoms. In solid state physics, a popular example of the interaction between noble gas atoms is the The Lennard Jones Potential . This interaction is clearly presented in "Introduction to Computer Simulation Methods: Applications to Physical Systems" by H. Gould and Jan Tobochnick 2nd edition as, 

V(r)=4*E*[(s/r)12 - (s/r)6 ]

with the force associated with this potential written as,

 F(r)=24*E*[2*(s/r)12 -(s/r)6] r

where E and s are the energy and distance parameter associated with this model.In the case of Argon, s=3.4x10-10 meters, E=1.65x10-12 Joules, m=6.69x10-26 kg. However, in this calculation, we use units of e_b=1x10-19 Joules, distance units of a_b=1x10-10 meters, mass units of m_b=1e-26 kg; so that, the time units are a_b*(m_b/e_b)1/2 . The simulation is done in one dimension. Finally, the simulation shows one atom fixed at the origin, another atoms sitting at an initial position. As the simulations is started, the molecules begin to interact as indicated by the motion of the right particle. The graph shown indicates the position of the moving particle as a function of time. Notice that the particle oscillates about an equilibrium point. This equilibrium point is actually the average distance between the particles known as the bond length. The simulation2 is capable of displaying the value of the particle's velocity as a function of time and the particle motion can be traced as well, all by checking the appropriate interactive buttons. The equation is solved by a Runge-Kutta solver using EJS.

2004 by Prentice-Hall, Inc. A Pearson Company