 # Mechanics

### Kinematics

If an object has position x1 at time t1 and position x2 at time t2, the average velocity during that interval is

v = (x2 - x1) / (t2 - t1).
On a graph of position vs. time, this should be familiar to you as the "rise over the run": the slope. So on a graph of position as a function of time, the velocity at a point on the path of an object's motion is the slope of the line at that time.

If the times t1 and t2 are very close, the average velocity becomes an instantaneous velocity. We will assume that we are working with instantaneous quantities in this book, unless otherwise noted. Note that if the position is decreasing (moving left on the real number line), the velocity will be negative: velocities depend on our choice of coordinate system.

We compute acceleration in exactly the same fashion as we did velocity. If an object has velocity v1 at time t1 and velocity v2 at time t2, the average acceleration during that interval is

a = (v2 - v1) / (t2 - t1).
On a graph of velocity vs. time, this is again the slope. So on a graph of velocity as a function of time, the acceleration at a point on the graph is the slope of the line there.
Acceleration can also be seen in a graph of position vs. time: at any point where the slope is not a constant, the graph is curved and the acceleration is nonzero. Since we restrict ourselves to situations involving constant acceleration (which includes constant zero acceleration!), we will see that when the graph is curved it will be parabolic, and the sign of the acceleration can be determined by the shape of the parabola: concave up corresponds to positive acceleration, and concave down corresponds to negative acceleration.
As before, if the times t1 and t2 are very close, the average acceleration becomes an instantaneous acceleration. If the velocity is decreasing, the acceleration will be negative; we do not usually use the term deceleration (except in the demonstration below).

### A Graphical Example

You need a Java-capable browser to be able to use the applets. If they do not work with your Windows system, download the Java VM (Virtual Machine) for your version of Windows at the download section at java.sun.com.

When you have decided on your answer, click on the "Check Answer" button to find out if you did the problem correctly. You may try the same type of problem with a different graph by clicking on the "Same Problem" button, or choose to practice with either the next or previous type in sequence ("Next Problem" or "Previous Problem" buttons), or a random type of problem ("Random Problem" button). These control buttons are a common feature on the applets in this text.

You need a Java-capable browser to be able to use the applets. If they do not work with your Windows system, download the Java VM (Virtual Machine) for your version of Windows at the download section at java.sun.com.

What kind of gadget could this applet be describing? Think of the toys often found in dentists' offices to help small children occupy their time in the waiting room: specifically the one with colored wooden balls that ride on wiggly wires. The graph in the problem describes a one-dimensional problem, not unlike any particular wire on one of those toys. Imagine a long straight wire with a red ball at some position. As a child moves it back and forth on the wire, the graph gives its position on the wire as a function of time. After a while, an adult puts a black ball on the wire at exactly the place the child started with the red one, and the child (miraculously!) moves it with precisely the same velocities and accelerations as she did with the first one.

3 * 9 * 9 * 9 * 9 * 9 * 9 * 9 * 9 * 9 * 9! / (2! 7!)

= 3 * 9 * 9 * 9 * 9 * 9 * 9 * 9 * 9 * 9 * 9 * 8 / 2

= 41,841,412,812

The "!" in the first expression denotes the factorial; n! is defined as the product of all the integers less than or equal to n, and 0! is defined as 1.

There are often combinatorial problems in physics. Our favorite is in particle physics. There, the problem is to add up the contributions from all possible particle interactions which can take place between the particle reactants and the particle products. The combinatorial problem there is interesting because many of the interactions are the same (since all particles of a given type are indistinguishable), and you are not allowed to count any of them more than once!

The next section continues with our discussion on kinematics as we construct kinematical equations.

©2008, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.

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