This problem you see in front of you is a
perfect example of where you might want to use
scientific notation because the numbers seem
so big when you look at them and the computation
might seem intimidating, especially in the
context of an exam if you’re trying to save time.
What we’re gonna do is try to solve this much
quickly and simply using scientific notation.
So what I’d like you to try to do is solve this
but use scientific notations instead and see
if you can speed up the process much more quickly.
If you try that, and pause, you can
give it another shot.
Let's jump in and see what it’s look like.
The first thing we’re wanna do if we’re gonna
solve this using scientific notation is rewrite
each of our three numbers in scientific notation.
So first, let’s rewrite this 3200 by again,
taking the decimal point from the right
and moving it one, two, three spaces over.
If this is the case then we have 3.2 times 10
to the third power, so this is the same number
just rewritten in scientific notation.
Doing the same thing with these other numbers,
We see, one, two, three, four, five, six places.
So this is 5 times 10 to the sixth power. And then
lastly, there’s a very very long number is
one two three four five six seven eight nine places.
And so this number, which we’re dividing is
times 10 to the ninth power.
To make this computation very easy, what we’re
going to do is not swipe over and take each of
these prefixes, the 3.2, the 5 and the 2.5
and bring them out front on their own.
3.2 times 5, divided by 2.5. And then we can make this
much easier just by putting all of our powers of 10
over here on the right so we have 10 to the third
times 10 to the sixth, divided by 10 to the ninth.
Now the last thing we have to do is be careful
to remember all of our rules of
multiplying exponents and they look like this.
10 to the three times 10 to the six is 10 to the,
just analyzing this part,
three plus six over 10 to the nine.
And you can see this will be 10 to the six plus
three which is nine, divided by 10 to the nine.
And so we see that these will cancel becoming one.
So this entire term on the right hand side
actually simplify down to being 1. Meaning that all
we have to do is to analyze these number out front.
3.2 times 5 divided by 2.5. A simple way to do this
might be to see how many times 2.5 goes into 5,
which is twice. Then this whole thing has
simplified down to 3.2 times 2, which is 6.4,
and this is our final answer, which looks
a lot simpler than the original problem worked
when we first jump into it. And this shows you
that scientific notation is a great way to
make any big problem be much more efficient just by
thinking about the powers of 10 out on their own.
Now we’re going to move to a… unit analysis and here’s
a great example of how you can think of unit analysis.
This equation you see in front of you is not something
that you should try to memorize right now
or worry about. The point is that we will in the future,
pretty soon, get into an equation, like this one,
which is an equation of motion,
where this x depends on many different variables.
X equals x for the zero plus v zero t and many other
things. And what we’re interested in is what if you were
in an exam setting and you didn’t remember exactly
what the equation was that you have in front of you,
whether it had an a times t in it, one half a times t,
or whether it’s one half a times t squared.
We can use unit analysis to quickly understand and
remember which of these it should be without having to
look back through any tables or any panicking.
Here’s how you do this.
If x depending on a times t squared versus x depending
a times t, we can look at the units of these.
These units are something that we’ll go over
in which you’ll know from a physical basis from
the problem and something that we’ll discuss. Specifically,
we’re gonna use brackets around the variables
like these square brackets around x, square brackets around t,
et cetera, to indicate the units of that variable.
So brackets around x means the units of x are,
and in this case the unit of x are meters.
This is indicating a position variable. The units of t
are seconds because this is our time variable
and that we’ll discuss soon. The units of a are meters
per second per second or meters per second squared.
If you know these things you can immediately
what we have above this information which is
does x depend on a times t squared or does x depend on
a times t. All we have to do to figure this out is
analyze the units. If I look at the units of a times
t squared, looking at these units below I can see that
the units of acceleration are meters per second squared.
So if I multiply by t squared,
I’m multiplying by seconds squared. So the seconds squared
from the numerator from t squared and the seconds squared
from the denominator from this variable a will cancel
evenly with just meters and that’s what we want,
is meters for this variable x. However if I look at this
quantity, a times t, it’s just meters per second squared
times seconds and only one of those units of second
will cancel, leaving me with incorrect units for x
which we know needs to be meters.
It will leave me with meters per second.
And so we can immediately see that the units of acceleration
times time, which is what this a is, acceleration,
the units of a times t are incorrect. They’re
per second. And so just by looking at the units
we can see that the units for this variable x,
if you’re writing this out in an equation,
must be a times t squared.