1. Nov 27, 2011

1. The problem statement, all variables and given/known data

I have an equation for motion:
x(t) = x(0) + x(0) * k * t$^{1.5}$
x is in meters and t is in seconds. I have to determine the unit of k.

2. Relevant equations

3. The attempt at a solution

x(t) = x(0) + x(0) * k * t$^{1.5}$
[x(t)] = meters
Therefore, [x(0)] = meters and [x(0) * k * t$^{1.5}$] = meters
meters * [k] * seconds$^{1.5}$ = meters
meters * (1/seconds$^{1.5}$) * seconds$^{1.5}$ = meters

Does this mean the the unit of k is 1/seconds$^{1.5}$, 1/seconds, or something completely different?

2. Nov 27, 2011

### technician

You first option is correct 1/s^1.5

3. Nov 27, 2011

Thanks, I didn't know if that was possible. I have another question if you could please check:

A force F is equal to k*x^n, where x is in centimeters.

[k*x^n] = N
[k]*(cm^n) = N
[k] = N/(cm^n)

The unit of k is N/(cm^n), is this correct?

Last edited: Nov 27, 2011
4. Nov 27, 2011

### technician

Yes, you seem to have it sorted out!!

5. Nov 27, 2011

Great, thanks. I've done far better in more difficult subjects like multivariable calculus, but the lack of a proper instructor for physics has me making stupid mistakes these days.

Would you happen to know of any online resource where I can practice graphical analysis of equations like these? What I have to do is to take a non-linear data set and convert it to a straight-line equation, determine appropriate units for slope and intercept, and determine values for the constants based on slope and intercepts. I only have two practice problems to work with.

6. Nov 27, 2011

### technician

I have never looked for any online resources for these types of problems.
If you know the power law for the equation (t^1.5 in your first example, n in your second example) Then the graph to plot is x against t^1.5 for the first and F against x^n for the second.
These would give straight lines with gradient k in each case.
If you do not know the power law.... I think that is the case in your second example, you only know it as n then you must take logs :
F = k * x^n
LnF = Lnk + n*Lnx

A graph of LnF against Lnx will be a straight line with gradient n and intercept Lnk from which k can be calculated

Does this make any sense for you, have you met log ~ log graphs

7. Nov 27, 2011