One Dimensional Motion:

We need to relate the three quantities position, velocity and acceleration. These quantities have both vector and scalar characteristics - click here for definitions of scalars and vectors. At the same time we will establish an important character of how physics is done. We start off with the simple case of constant acceleration and see what velocity and position result. It's best to do this graphically:

	Here the acceleration (the rate of change of velocity) is constant at 2 m/s. This means that for every second that goes by the velocity increases by 2 m/s. After 1 second the velocity is 2 m/s; after 2 seconds it is 4 m/s, etc. This is easy to figure out in your head.
	This is a plot of the velocity we got from the constant acceleration above. It increases linearly with time.
	How do we calculate the distance covered (i.e. displacement or, since we start at x = 0, position)? Well, distance = rate (i.e. velocity) x time - remember that old formula? Here we have to be careful - the velocity changes with time, as the curve above shows. How to handle that? Take the average velocity. During the first second the velocity starts at 0 and ends at 2 m/s. The average is (0+2)/2 = 1 m/s. 1 m/s for 1 second gives a distance of 1 meter covered in the first second. Do the same for each succeeding second - it's easier if you start from the beginning each time. For example, after the 2nd second the average velocity is (0+4)/2 = 2 m/s. 2 m/s for 2 seconds provides x = 4 m. The result is the curve at the left. You should confirm this result.

OK - this is easy enough, but the real value of this exercise is this: We verified the curves by using what we know about velocity, acceleration and position, and how they are related. For the simple case here, this was fairly easy. Look at this from a different viewpoint: Consider only the geometry of the curve a = 2 m/s² .  It is essentially a rectangle of height 2 m/s² and width t (t can take on any value - in these pictures between 0 and 5 seconds). What is the area accumulated under this curve during the first second? It's simply width times height, or 2 x 1 = 2. This is precisely the value of velocity after the first second, and if the product of the units is considered (m/s² x s), the units of the result are those of velocity! We get the correct value, including units, without having to consider the actual physical problem at all! Moral: Mathematics is truly the language of physics.

So now we have it: The area under the acceleration curve is a measure of velocity. You should verify for yourself that the area under the velocity curve is a measure of position. And this holds for any acceleration curve, no matter how complex. But it gets better: Measure the slope of the velocity curve: Rise over run - you remember that. What is it? Well, since the slope is constant, any rise/run ratio will do, say 10 m/s rise divided by the corresponding (from the curve) 5 second run. The result is 10 m/s divided by 5 seconds = 2 m/s² - the constant acceleration.

Summary: The area under the acceleration curve is the velocity; the area under the velocity curve is position. The slope of the position curve is velocity; the slope of the velocity curve is acceleration. And it's all math - the physics comes along for free.

Question: how do we figure these slopes and areas if the curves get messy?