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I would like to know if an acceleration number would remain squared in

$$ v=v_{o}+at $$

Such as 1.35 m/s^2, for example, would end as

$$ v=v_{o}+(1.35^2)t $$

or simply as

$$ v=v_{o}+(1.35)t $$

Thank you very much for any help!

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  • $\begingroup$ No arguments with the answers below, but your intuition about the square being somewhere in the math is a good one. If write acceleration as a function of variables representing each of the base units, then the variables for distance and time would have to appear as they do in the unit for acceleration that is, m/s^2 must represent a value that, at some level has distance divided by time squared. (ex: a = 2 d / t^2, if the initial velocity is zero). Of course it can be more complicated than this simple example, but the fundamental idea is sound. $\endgroup$ Jan 9 at 21:08
  • $\begingroup$ jvno, note that all units on both sides of the equal sign must match, or the equation is incorrect. And as Jason Patterson pointed out, ALL numbers and variables have units associated with them unless you are dealing with a dimensionless form (the kinematic equations are NOT dimensionless). $\endgroup$ Jan 10 at 0:19

6 Answers 6

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No it would not be squared. Here $a$ is in $m/s^2$ as you rightly said and $t$ is in $s$ such that $at$ is in $m/s^2\times s=m/s$ and has the right unit of velocity by using the value of 1.35. (That is, the correct expression is $v=v_0+(1.35)t$).

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    $\begingroup$ This is sloppy language. The correct expression is $v=v_0+(1.35{\:\rm m/s^2})t$. $\endgroup$ Jan 9 at 23:14
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The value 1.35 is just the value of the acceleration. No need to square it. You only notice squared units when you convert between different units. For example:

$$5\,\text{cm}^2=5\times (\text{cm})\times(\text{cm})=5\times(10\text{mm)}\times(10\text{mm)}=500\,\text{mm}^2$$

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$1.35 \,\rm m/s^2$ means $\underbrace{1.35}_{\rm number}\times \underbrace{1 \,\rm m/s^2}_{\rm unit\,value}$ (an SI unit of acceleration) just as $10\,\rm m$ means
$\underbrace{1\,\rm m + 1\,\rm m+ 1\,\rm m+ 1\,\rm m+ 1\,\rm m+ 1\,\rm m+ 1\,\rm m+ 1\,\rm m+ 1\,\rm m+ 1\,\rm m}_{10\times 1\,\rm m}$

Thus $a \times 1 \,\rm m/s^2 = 1.35 \times 1\,\rm m/s^2\Rightarrow a = 1.35$.

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You should treat units as some constants. E.g. if you have an acceleration of 1.35 meters per second per second, it just means that you take a constant that denotes the meter (call it $\mathrm{m}$), and another constant that denotes a second (call it $\mathrm{s}$), and you multiply these values raised to the appropriate powers:

$$a=1.35\cdot\mathrm{m}\cdot\mathrm{s}^{-2}.$$

Then, when inserting this value into an expression like e.g. $\Delta v=at,$ you take all the appropriate quantities in a similar form, e.g. if $t$ is five seconds, then you get

$$t=5\cdot\mathrm{s},$$

which you can then substitute to get

\begin{align} \Delta v=at&=1.35\cdot\mathrm{m}\cdot\mathrm{s}^{-2}\cdot 5\cdot\mathrm{s}=\\ &=6.75\cdot\mathrm{m}\cdot\mathrm{s}^{-1}=\\ &=6.75\,\frac{\mathrm{m}}{\mathrm{s}}. \end{align}

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Such as 1.35 m/s^2, for example, would end as $$ v=v_{o}+(1.35^2)t $$ or simply as $$ v=v_{o}+(1.35)t $$

Neither of those expressions make sense as written, because the units would be inconsistent. You can see that if you plug in some arbitrary values, for example $v_o = 36\ \mathrm{km/h}$ and $t = 0.37\ \mathrm{min}$. (I deliberately chose those to use different time units.) Then your $v$ would come out to e.g.:

$$\begin{aligned} v &= 36\ \mathrm{km/h} + 1.35 \cdot 0.37\ \mathrm{min} \\ &= 36\ \mathrm{km/h} + 0.4995\ \mathrm{min} \\ &= 10\ \mathrm{m/s} + 29.97\ \mathrm{s} \\ &= \text{nonsense!} \end{aligned}$$

and there's nothing you can do to $10\ \mathrm{m/s}$ and $29.97\ \mathrm{s}$ to get them to have the same units, which means you cannot add them together and get a physically meaningful value.


The correct solution, instead, is to include the units of your acceleration $a$ (in this case $a = 1.35\ \mathrm{m/s^2}$) in the expression, like this:

$$v = v_o + (1.35\ \mathrm{m/s^2})\,t$$

Then, plugging in the rest of the values with their respective units, you'll get a dimensionally consistent expression:

$$\begin{aligned} v &= 36\ \mathrm{km/h} + (1.35\ \mathrm{m/s^2}) \cdot 0.37\ \mathrm{min} \\ &= 36\ \mathrm{km/h} + 0.4995\ (\mathrm{min \cdot m/s^2}) \\ &= 10\ \mathrm{m/s} + 29.97\ \mathrm{m/s} \\ &= 39.97\ \mathrm{m/s} \\ &≈ 40\ \mathrm{m/s} \end{aligned}$$

(Huh, I didn't actually expect it to work out so close to such a nice round number when I picked my random values.)


BTW, as others have noted, when calculating with units you should treat each unit as if it were an (unknown) multiplicative constant. In other words:

$$1.35\ \mathrm{m/s^2} = 1.35 \cdot (1\ \mathrm{m}) \cdot (1/\mathrm{s^2})$$

The one case where "treating units as multiplicative constants" does not work is with "affine units" such as degrees Celsius or degrees Fahrenheit that conceal an extra additive constant in the notation. If you need to deal with those, the general solution is to convert them into the corresponding non-affine units such as Kelvins first.


Now, in the bad old days in the 1980s and '90s, back when we were using primitive calculators that didn't understand physical units, the way you'd actually calculate this result would be to first convert all the values to matching units (i.e. all the distances to meters, all the times to seconds, etc.), like this:

$$\begin{aligned} v &= 36\ \mathrm{km/h} + (1.35\ \mathrm{m/s^2}) \cdot 0.37\ \mathrm{min} \\ &= 10\ \mathrm{m/s} + (1.35\ \mathrm{m/s^2}) \cdot 22.2\ \mathrm{s} \end{aligned}$$

and then, after carefully verifying that the units really matched, you'd "drop the units" and use your calculator to calculate e.g.:

$$v_{\rm unitless} = 10 + 1.35 \cdot 22.2 = 39.97$$

to which you'd again append the correct (manually determined!) units (in this case $\mathrm{m/s}$) to arrive at $v = 39.97\ \mathrm{m/s} ≈ 40\ \mathrm{m/s}$.

You could justify this "dropping the units" with some algebra: basically you'd need to algebraically rearrange the expression to group all the numbers together and the units together, like this:

$$\begin{aligned} v &= 10\ \mathrm{m/s} + (1.35\ \mathrm{m/s^2}) \cdot 22.2\ \mathrm{s} \\ &= 10\ \mathrm{m/s} + (1.35 \cdot 22.2)\ (\mathrm{m/s^2 \cdot s}) & \text{(commutative law)}\\ &= 10\ \mathrm{m/s} + (1.35 \cdot 22.2)\ \mathrm{m/s} & \text{(cancel 1/s and s)} \\ &= (10 + 1.35 \cdot 22.2)\ \mathrm{m/s} & \text{(distributive law)} \end{aligned}$$

and then use the calculator to evaluate the unitless expression inside the parentheses. Few people would ever actually write all these steps out, but in principle you could do it.

Even so, obviously this was a very error-prone process, since there was nothing (except your own eyes and brain) to check that you weren't accidentally adding together numbers with different units and getting nonsense as the answer.


Fortunately, however, we live in the 21st century now and everyone (with a working Internet connection, at least) has access to a calculator with full support for all commonly used physical units right at their fingertips.

Just go to Google and type e.g. "(36 km/h) + (1.35 m/s^2) * (0.37 min)" into the search box and press enter (or just type it directly into your browser's address bar!). You'll get your answer right back:

Screenshot of "(36 km/h) + (1.35 m/s^2) * (0.37 min) = 39.97 m/s" calculated using Google

There are also plenty of offline apps and software tools and even dedicated "scientific calculators" that can handle physical units, convert between them and let you know if your expression has inconsistent units. But honestly, 99% of the time when I need to do a simple physics calculation, I just use Google. It's there, it works, why would I bother with something else?

(The one exception is when, as a programmer, I need to write numerical code in an old-fashioned programming language like C or Java or even JS that doesn't natively understand physical units or have convenient syntax for them. In those cases I do need to resort to the old "unitless" method and just test my code carefully to make sure I don't make any unit mistakes. At least more modern languages are starting to support physical units better, although e.g. with Python the problem is more choosing which third-party package to use…)

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    $\begingroup$ The Google Calculator is great. It has lots of built-in constants, eg particle & planet masses. But it's not perfect: eg, it uses the tropical year for the light-year instead of the Julian year, which has been the IAU standard for 4 decades. en.wikipedia.org/wiki/Light-year (365.25 * 86400 seconds) * c - (1 lightyear) $\endgroup$
    – PM 2Ring
    Jan 10 at 2:48
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The other answers are correct you don't square the number but to give an intuitive sense of why they are right; it's because the number is already a product.

Imagine the area of a 2 meter by 4 meter area.

2 m * 4 m

= (2 * 4 ) * ( m * m )

= 8 m^2

From that you can see it doesn't make sense to additionally square the 8 because it was already the product of 2 * 4

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  • $\begingroup$ That's a nice example, although I don't think it holds generally. You can have an acre of land without being required to decompose it into a product of two lengths (indeed the acre could be in some odd, non-rectangular shape). To me the square in $m^2$ has more to do with the scaling properties of area if you change units of length. $\endgroup$
    – Andrew
    Jan 9 at 22:51
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    $\begingroup$ @Andrew, an area of any shape is a sum of products, possibly an infinite sum obtained using calculus. $\endgroup$
    – Peter
    Jan 10 at 0:53

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