Treating units algebraically and dimensional analysis | Algebra I | Khan Academy
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Treating units algebraically and dimensional analysis | Algebra I | Khan Academy

Voiceover:We’ve seen
multiple times in our life that distance can be
viewed as rate times time. What I want to do in this video is use this fairly simple
formula right over here, this fairly simple equation, to understand that units
can really be viewed as algebraic objects, that you can treat them like variables as we work through a
formula or an equation, which could be really, really helpful to make sure that we’re
getting the results in units that actually make sense. For example, if someone
were to give you a rate, if they were to say a rate of, let’s say, 5 meters per second, and they were to give you a time, a time of 10 seconds, then we can pretty, in a
pretty straightforward way, apply this formula. We say, well, distance
is equal to our rate, 5 meters per second times our time, times our time, which is 10 seconds. What’s neat here is we
can treat the units, as I’ve just said, like
algebraic constructs, kind of like variables, so this would be equal to, well, multiplication, it doesn’t matter what order we multiply in, so we can change the order. This is the same thing as 5 times 10, 5 times 10 times meters per second, times meters per second times seconds, times seconds. If we were to treat our units as these algebraic objects, we could say, hey, look, we have seconds divided by seconds, or you’re going to have
seconds in the denominator multiplied by seconds in the numerator. Those are going to cancel out, and 5 times 10, of course, is, 5 times 10, of course, is 50. We would be left with 50, and the units that we’re
left with are the meters, 50 meters. That’s pretty neat. The units worked out. When we treated the units
out like algebraic objects, they worked out so that
our end units for distance were in meters, which
is a unit of distance. Now you’re saying, “OK,
that’s cute and everything, “but this seems like a little
bit of too much overhead “to worry about when I’m just doing “a simple formula like this.” But what I want to show you is that even with a simple formula like distance is equal to rate times time, what I just did could
actually be quite useful, and this thing that I’m
doing is actually called dimensional analysis. It’s useful for something as simple as distance equals rate times time, but as you go into physics
and chemistry and engineering, you’ll see much, much, much more, I would say, hairy formulas. When you do the dimensional analysis, it makes sure that the
math is working out right. It makes sure that you’re
getting the right units. But even with this, let’s try a slightly
more complicated example. Let’s say that our rate is, let’s say, let’s keep our
rate at 5 meters per second, but let’s say that
someone gave us the time. Instead of giving it in
seconds, they give it in hours, so they say the time is equal to 1 hour. Now let’s try to apply this formula. We’re going to get distance is
equal to 5 meters per second, 5 meters per second times
time, which is 1 hour, times 1 hour. What’s that going to give us? The 5 times the 1, so we multiply the 5 times the 1, that’s just going to give us 5. But then remember, we have to treat the units algebraically. We’re going to do our
dimensional analysis, so it’s 5, so we have meters per second times hours, times hours, or you could say 5 meter hours per second. Well, this doesn’t look like a … This isn’t a set of units that we know that makes sense to us. This doesn’t feel like our
traditional units of distance, so we want to cancel this out in some way. It might jump out of you, well, if we can get rid of this hours, if we can express it in terms of seconds, then that would cancel here, and we’d be left with just the meters, which is a unit of distance
that we’re familiar with. So how do we do that? We’d want to multiply this thing by something that has
hours in the denominator and seconds in the numerator, times essentially seconds per hour. How many seconds are there per hour? Well, there are 3,600 … Let me do this in a … I’ll do it in this color. There are 3,600 seconds per hour, or you could even say that there are 3,600 seconds for every 1 hour. Now when you multiply, these hours will cancel with these hours, these seconds will cancel
with those seconds, and we are left with, we are left with 5 times 3,600. What is that? That’s 5 times 3,000 would be 15,000, 5 times 600 is another 3,000, so that is equal to 18,000. The only units that we’re left with, we just have the meters there. 18- Oh, it’s 18,000, 18,000, 18,000 meters. We’re done. We’ve now expressed our distance in terms of units that we recognize. If you go 5 meters per second for 1 hour, you will go 18,000 meters. But let’s just use our little dimensional analysis
muscles a little bit more. What if we didn’t want
the answer in meters but we wanted the answer in kilometers? What could we do? Well, we could take that 18,000 meters, 18,000 meters, and if we could multiply it by something that has meters in the denominator, meters in the denominator and kilometers in the numerator, then these meters would cancel out, and we’d be left with the kilometers. So what can we multiply it so we’re not really changing the value? We want to multiply it by essentially 1, so we want to write equivalent things in the numerator and the denominator. So 1 kilometer is equivalent to, equivalent to 1,000 meters. One way to think about it, we’re just multiplying this thing by 1, 1 kilometer over 1,000 meters. Well, 1 kilometer is 1,000 meters, so this thing is equivalent to 1. But what’s neat is when you multiply, we have meters canceling with meters, and so you’re left with
18,000 divided by 1,000 is equal to 18. And then the only units we’re left with is the kilometers, and we are done. We have re-expressed our distance instead of in meters in terms of kilometers.

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16 thoughts on “Treating units algebraically and dimensional analysis | Algebra I | Khan Academy

  1. I had understood the process on how to get the answer but confused on why at 4:12 do you have to write 3600 sec over 1 hr to get the answer when simply just have timesed 5 m/s by 3600 s.

  2. Sal doesn't write all the steps of the algebra here (but he does narrate the process). I went through the whole KA math curriculum up to this point (preparing for university at the age of 38) and I do understand this video. It actually shows an easier way of unit conversion than what was encountered earlier in the curriculum. I was struggling before a bit.

    To understand what's going on, you need to know these topics well:

    1. fractions
    You gotta master fractions. They pop up everywhere and they aren't that intuitive in the beginning. After some time they become pretty easy, like basic addition. Trust me. How to master them? Solve problems until you feel sure about them.

    2. ratios, rates, proportions (
    This topic is based on fractions, so if you don't know fractions well, you will SUFFER. Master fractions first. This was the first math topic where I had to stop for several days and process it. It wasn't intuitive at all. I watched the videos again, I analyzed the problems presented and solved practice problems. It eventually became a relatively comfortable topic but my intuition wasn't fully developed. However, I knew that this topic was going to return in the future and here I am 🙂 As I said, now I understand this video quite easily.

    P.S. All that letter (m, s, h…) cancellation woodoo is actually just simple fractions process (2/2 = 1, m/m = 1). Where it gets tricky is rates (m/s vs. m/h v.s km/h…). If you don't know how fractions work and you don't know how a rate (e.g. meters per second) is "made", you will have a very very hard time understanding it from this video.

  3. please help me find the dimensions of k in the formula of workdone on an external agent given by the formula W=-kx^2

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