# Can you simplify dimensional analysis questions before plugging in values?

If I were to solve the equation:

$$d=\frac{a^3}{cb^2}$$

and values were given as so: $a=9.7\ \mathrm m$, $b=4.2\ \mathrm s$, and $c=69\ \mathrm{m/s}$.

Why would it be wrong for me to use dimensional analysis first?

Using dimensional analysis you would simplify: $\mathrm{m^3/((m/s)(s^2))}$ to $\mathrm{m^2/s}$.

That is right as far as I know, but when I plug in the values I get $d=.07\ \mathrm{m^2/s}$, that is I plug it in as $(9.7\ \mathrm m^2)/(69\ 4.2\ \mathrm s)$ since I already did dimensional analysis.

However, when I plug in the values as $((9.7^3)(\mathrm m^3))/((69\mathrm m/\mathrm s)((4.2^2)(\mathrm s^2)$ I get the right answer of $d=.75\ \mathrm{m^2/s}$.

I know my dimensional analysis is correct but why is it that I cannot simplify the original equation before plugging in the values (as in I have to plug it in as $9.7^3\ \mathrm m^3$ instead of $9.7\ \mathrm m^2$)?

So essentially in this question I should think of it as for example if I take the equation $5x^2$, then $5x^2$ is equal to $25x^2$? Is that how I should think of it?

• The exponents apply both to the units and to the numbers. – Javier Jan 5 '18 at 19:49
• Oh right, I did forget about that rule, but can you explain why the exponent applies to both the units and numbers (specifically why does it apply to the unit)? – dawnwall Jan 5 '18 at 19:51
• Imagine if you were to evaluate (ab)^2, where a is a constant representing the value or magnitude and b is representing the unit. You would get a^2 * b^2 instead of just a*b^2 (like what you have shown in your question). – Kane Billiot Jan 5 '18 at 23:02
• That makes a lot of sense. I just want to clarify for the example given (ab)^2 it follows the power of a product rule, and in my question it does not. cb^2 would just be c*b^2 and as with a^3 you apply the product rule with the units, that is it becomes: 9.7^3(m^3) according to the power of a product rule. Thanks so much that was helpful – dawnwall Jan 6 '18 at 2:57