This is from Strang's Calculus book:
When $f$ is in feet and $t$ is in seconds what are the units of $f'$ and its derivative $f''?$
In $$f=16t - 16t^2$$ the first $16$ is ft/sec but the second $16$ ______
I remember from physics that the second derivative is acceleration and the dimensions are $\text{ft/s}^2,$ but how do I deduce it from the formulas?
What about the second "$16$" in the formula above? Can't make a clue about it.
On
The units must be consistent. All the quantities that you add or subtract on the left hand side and right hand side have the same units. Since $f$ is in ft, that means that $16t$ is also in ft. and that means that the units of the first $16$, when multiplied with s, must be equal to ft, so the unit is ft/s. Similar reasoning for the second term means that the units are ft/s^2.
On
Notation: we say that the units of variable $y$ are $[y]$. So it is a little-known-in-calculus-textbooks fact that the units of $$\left[\frac{dy}{dx}\right]=\frac{[y]}{[x]},$$ and the units of $$\left[\int y\,dx\right]=[y]\,[x].$$ This implies that the units of $$\left[\frac{df}{dt}\right]=\frac{[f]}{[t]}=\frac{\text{ft}}{\text{s}},$$ and $$\left[\frac{d^2f}{dt^2}\right]=\frac{[f]}{[t]^2}=\frac{\text{ft}}{\text{s}^2}.$$
As for going from your formulas, the two $16$'s have hidden units associated with them to make the final units work out. The rule is that you can only add identical units. The first $16$ evidently has units of $\text{ft}/\text{s},$ so as to cancel out the s from the $t,$ and the second $16$ evidently has units of $\text{ft/s}^2,$ to cancel out the units from the $t^2.$
When taking the first derivative $\frac{dx}{dt}$, we remark that we get a measure of time on the "denominator" of the derivative. The same goes for $\frac{d}{dt}\frac{dx}{dt}$. We get a supplementary measure of time in the "denominator", which provides an intuition as to why the dimensions of acceleration are $\frac{L}{T^2}$. The number 16 does not change the dimensions of any quantities in the function since it is a scalar.