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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?

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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.

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  • $\begingroup$ But I thought that constants never have units? $\endgroup$ – fYre Sep 25 '13 at 2:49
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    $\begingroup$ @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
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    $\begingroup$ 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
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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, $L/T^2$).

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  • $\begingroup$ 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
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    $\begingroup$ @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

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