Interesting Link:  Some Berkeley Physics Demonstrations

Introduction:

There are two graphs that are often used: position vs time and velocity vs time. Since we will consider only one-dimensional motion, the direction of a vector can be completely specified using an algebraic sign.

There are three quantities which can be obtained from the graphs: distance/displacement, speed/velocity, and acceleration. Distance is a scalar and displacement is a vector, but distance is not necessarily the magnitude of the displacement vector. For example, a car that travels one mile north and then one mile south has traveled a distance of two miles. But the displacement is defined as the vector from the origin to the final position and it is the null vector! So the magnitude of the displacement is zero; it is not the distance in this case!

Average speed is defined:

Average velocity is:

where s is the symbol usually used for general displacements in any direction. For one-dimensional motion, we use x. Average acceleration is:

Average Velocity:

Suppose we have a rather strange go-cart:
 

During the first two seconds, it travels 10 meters forward. Since the slope is constant, the average velocity is the same as the instantaneous velocity at any time between zero and two seconds:

Because the motion is in one direction and is constant, the speed is the magnitude of the velocity:

But at 2 seconds, the velocity changes abruptly in magnitude and direction to -2.0i m/s; the average velocity from 0 to 4 sec is NOT the arithmetic average of 5i and -2.0i which would be 1.5i m/s; it is:

(A change in any measured quantity is the final value minus the initial value.) Can you find the average velocities from 4 to 7 seconds and over the entire interval? After you've tried it, here are the answers.

Displacement vs Distance; Instantaneous Velocity and Acceleration:

No real go-cart could change velocity abruptly, but that's what this one does! The plot below uses the slopes from the graph above.

The slope of the velocity vs time graph is the acceleration. All of these accelerations are zero, because each velocity is constant within its time interval. We'll look at a non-zero acceleration a little later.

The signed area under the v vs t graph is the displacement. The total distance traveled is the sum of the unsigned areas.

Here are the displacements and distances for several time intervals. Since this is one-dimensional motion, a sign indicates the direction of the displacement vector:
 

What is the displacement and the distance from 4 to 8 seconds? After you've worked it out, here's the solution.

Distance traveled can also be obtained from the position vs time graph if you're careful.

Example: Suppose a car travels from Albany, NY to New York City (NYC). I don't know what the distance is by the interstate highway, but let's assume it's 180 miles one way and we'll also assume that it's straight (to make it one-dimensional). The displacement is 180 mi south, while the distance is 180 m. If the car were to travel from Albany to NYC and back, the displacement would be the null vector, but the distance would be 360 miles. Suppose the car continues north to Montreal, Canada (assume another 240 mi). The car's displacement would be 240 mi north but the distance traveled would be 600 mi. Taking south as positive, the origin at Albany, and a constant 60 mph, the graph is:

Now suppose we have a displacement-time graph for an even stranger go-cart:

Can you see that the average velocity from 1.5 to 4 seconds is (you might estimate slightly different values from the graph):

Notice that this is the slope of the chord joining the two points on the graph? Now you try it. Find the average velocity from 2 to 4 seconds, and from 2 to 6 seconds. Here are the solutions.

While the average velocity is the slope of the chord, the instantaneous velocity is the slope of the tangent line. For example, the instantaneous velocity at t = 4 seconds is the null vector, since the tangent drawn to the curve at 4 seconds is horizontal. To find the instantaneous velocity at t = 2 seconds, we have to "eyeball" a tangent line as shown below:

The slope of the tangent is

You might get a slightly different value, depending on how you draw the tangent.

When the equation of the curve is known, the derivative can be used to find the instantaneous velocity: The equation is (because I used this equation to set up the data):

Since the derivative is the slope of the tangent, dx/dt = v:

The substitution of t = 2 seconds yields a velocity of -4 m/s. My "eyeball" guess at the tangent line was not bad!

Acceleration vs Time Graphs:

You've seen that "area" need not be in square meters (for example). The area under the velocity-time graph has units of meters, since it is obtained from the product of m/s and s. Likewise, the area under the acceleration-time graph has units of (m/s²)(s) = m/s or velocity. In figure P2.19 on page 47 of Serway, where a graph of acceleration vs time is given, the area under the line from 0 to 5 sec is (2 m/s²)(5 s) = 10 m/s. From 0 to 10 sec, it is 20 m/s, showing that the velocity increases with the area taken. (This makes sense, since the velocity would increase when the object is under acceleration.)