Just to make sure we agree on what is going on with the math under discussion, and because I haven't posted anything for a few days, I will do a quick review of the proper math notation.
Given: X = a * b = ab
This means that the quantity X is equal to a multiplied by b. The 2 forms are equivalent.
Given: X = a * b^n = ab^n
This means that the quantity X is equal to a multiplied by b raised to the n power. Again the 2 forms are equivalent.
Given: X = a^n * b^n = (ab)^n
This means either that a is raised to the n power and b is raised to the n power and then the 2 results are multiplied or a is multiplied by b and the result is raised to the n power. The 2 forms provide the same result.
Notice that there are in fact more ways to indicate raising a value to a power. This is what I have found to be one of the more common ways. It comes from computer programming where a common way to write an equation to be evaluated by the program is to use the ^n form. In fact if you provide appropriately formatted variables in a program the statements above would be successfully performed by many programming languages.
And in regard to how to handle units. When performing mathematical operations on units of measure such as meters (m), grams (g), Joules (J), etc the proper way is to use the same mathematical operations on the units as on the metrics that go with them. For Example:
Acceleration has the units of m/s/s = m/s^2 (meters per second squared). That is meters divided by time squared. Notice that the ^2 applies only to s, if I wanted it to apply to m/s I would need to contain (m/s) in parentheses. So m/s^2 is quite different from (m/s)^2. The first way squares s the second way squares both m and s.
The way that things work out you can determine the speed that something will attain under acceleration by the formula:
v=at Where v is velocity, a is acceleration and t is the time the acceleration takes place.
Example:
Given a = 9.80 m/s^2 (9.80 meters per second squared)
This is approximately the acceleration of a falling object at the Earths surface.
What is the velocity of a falling object after 10 seconds?
Acceleration has the units m/sec^2.
So: v = (9.80m/s^2) * 10s
Notice that I enclosed the 9.80/^2 in parentheses to indicate that the 10s is not a multiplier of s^2, but is applied to the whole expression.
Now rearrange it a little bit to separate the units from the metric.
v = 9.80 * (m/s^2) * 10 * s
10 multiplied by 9.80 = 98
(m/s^2) multiplied by s = m/2
Therefore: v = 98m/s
And that is how you handle units in science.
Bill Gill
