Form what I understand if you have an equation such as:

$$v = v_0 + at$$

then the dimensions must match on both sides i.e. $L/T = L/T$ (which is true in this case), but I have seen equations such as 'position as a function of time' $x(t) = 1 + t^2$, and obviously time is in $T$, but apparently the function gives you position which is $L$... so what happens to $T$ and where does the $L$ come from? I thought dimensions must always match...

Also, let us say that you know the time to reach a destination is proportional to distance i.e. double the distance and you get double the time, now this makes sense to me, but as I said earlier I thought that dimensions must always be consistent or else you can not make comparisons in physics, so if you are giving me $L$ (the distance), how can that become $T$ (time) all of a sudden?


Units must always be consistent, that is correct. So using your example of:

$$ x(t) = 1 + t^2 $$

where the left hand side has units of $L$ (distance). This means the constant $1$ on the right side has implied units of $L$ while the coefficient in front of $t^2$ (which has the value of $1$) has implied units of $L/T^2$.

In other words, the units do match but they get attached to the constants multiplying each term.

| cite | improve this answer | |
  • $\begingroup$ But I thought that constants never have units? $\endgroup$ – fYre Sep 25 '13 at 2:49
  • 1
    $\begingroup$ @HaniSayegh They must if you need them to balance out the units of the equation. In your example, $x_0 = 1$ which is why the $1$ appears. So you can certainly have constant values of something that has units. $\endgroup$ – tpg2114 Sep 25 '13 at 2:51
  • $\begingroup$ Likewise, $a = 2$. So there are units attached to that also. $\endgroup$ – tpg2114 Sep 25 '13 at 2:52
  • $\begingroup$ Let us say you are given the following instead: x(final) = 2x(initial), then the '2' would be dimensionless in this case right? $\endgroup$ – fYre Sep 25 '13 at 2:53
  • 2
    $\begingroup$ Constants don't have units? How about g = 9.8 m/s2 ? c = 300.000.000 m/s ? $\endgroup$ – MSalters Sep 25 '13 at 7:43

Write the function $x(t)$ in the form $$ x(t) = x_0 + v_0t + \frac12a_0t^2 $$ In the case you have given, then $x_0=1$, $v_0=0$ and $a_0=2$. Since $[v_0]=L/T$ and $[a_0]=L/T^2$, then the units match: $$ [x(t)] = [x_0] + [v_0][t] + \frac12[a_0][t^2] = L+\frac{L}{T}T+\frac{L}{T^2}T^2=L $$ The dimensions do indeed match, you are merely forgetting that the coefficients in front of $t$ (i.e., 1) have units as well (in this case, $L/T^2$).

| cite | improve this answer | |
  • $\begingroup$ Assume we were give x(t) = t^3,then what equation would you use to get L? $\endgroup$ – fYre Sep 25 '13 at 3:15
  • 5
    $\begingroup$ @Hani the point of this answer is that technically, $x(t) = t^3$ is wrong. In practice, it's shorthand for $x(t) = Ct^3$, where $C$ is some constant (of dimension $L/T^3$) that you should be able to infer from the context. $\endgroup$ – David Z Sep 25 '13 at 4:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.