Basic Question About Canceling Units This is a very basic question but, here it is. In a given equation, where I have a term such as 4 m/s and I multiply this by the variable t, which is time, why is it incorrect for me to cancel out t and s? It is because s is a constant and t is a variable?
 A: You cannot cancel it because the s in 4 m/s is a unit (seconds) and t is a variable (time), which is measured in seconds. So the units cancel, but not the numeric value. Assume $t = 2s$, then you get $4 m/s \cdot 2 s = 8 m$.
A: You need to think in terms of quantities. It looks like you are using some equation for distance without acceleration:
$$x-x_0=vt$$
where $x$ and $x_0$ are quantities called "position"  (specifically at time $t$ and $0$) and will have units of distance. $v$ is a quantity called "velocity" and has unit of distance/time, and $t$ is a quantity called "time" and has unit of time.
Let's say that $x_0 = 0$m, $v=4 $m/s, and $t=1$ min. Then you would have
$$x=4~\mathrm{m\cdot min/s}.$$
While that seems weird, that is a correct position quantity. It, however, is not a standard distance unit. The unit "min/s" itself can be viewed as a dimensionless number of 60:
$$x=4\cdot 60~\mathrm m = 240 \mathrm m$$.
So, no, $t$, a quantity, cannot cancel $s$, a unit. You must remember that quantities have units, and matching units can cancel.
An example in astrophysics of weird mixed units is used with the Hubble constant:
$$H_0\simeq 69.4 \mathrm{\frac{(km/s)}{Mpc}}$$
km (kilometer) and Mpc (megaparsec) are both distance units, so $H_0$ actually could be expressed in inverse seconds (and some people might actually use hertz (arrrgh) ), but it would be a terribly small number, and the units used above express an important concept: how does the expansion speed behave for different separation distances.
A: Essentially you are right in your last statement.
Let's define $t_{1s} \equiv 1\ {\rm s}$. Then your question amounts to asking whether this equation is true for any $t$:
\begin{equation}
\frac{t}{t_{1s}} \stackrel{?}{=}1
\end{equation}
The answer is, no, this equation is not always true. It is only true if $t=1{\rm s}$.
A: $t$ contains both a unit AND a number. You can't use a unit to cancel both a number and a unit. You don't know what unit you are going to put into the variable ahead of time (seconds vs hours, meters vs km, etc).
And even if you did know, you can't cancel only the unit part of $t$ while leaving the number part of $t$ in the equation. That would leave you with some other variable to replace $t$ except it has no units but paradoxially must be the replaced with a numerical value measured in the units you just cancelled out. That doesn't make sense (and if someone else walks up to your equation they won't know what those units are because they're gone)
So best to leave the variable untouched until you actually replace it with its constintuent parts: a number with units. Then you can cancel the units and the number remains behind.
