Leah here from leah4sci.com/MCAT, and in this video, we'll talk about Vectors and finding
vector components. In part 1 of this series we talked about difference between a Vector
and Scaler and said that Vectors have a magnitude and a direction. But when you're looking at
the direction of a vector, the direction of the vector can be an x-direction, it can be
in the Y-direction or it can be in some direction between the x and y axis. If you're trying
to understand and calculate the x and y components of this non-traditional line, you have to
know how to break up a vector into its x and y components. For example, you may be faced
with a question like this: A bird takes off from the ground at a 30-degree
angle and lands on a tree branch exactly 50 meters away. How far did the bird fly with
respect to the ground? How high is the branch? The bird flew exactly 50 meters, but because
the bird flew at an angle, the distance on the ground, meaning if you were to follow
under the bird as it flew, would not be 50 meters. Let's represent this with the diagram.
The ground will be our x-axis and the bird takes off at 30 degrees so our angle is 30
for a total distance of 50 meters. If the bird line on the tree over here, if
we're following along the ground, we would have walk something less than 50 meters and
if we want to know the height of the branch that will be significantly less than 50 meters.
Now even though this isn't draw on the scale you can get some idea that the x distance
will be closer to 50 than the y. So how do we figure out the x and y components?
We'll have to use the equation for finding the x and y components of a vector as follows:
If we call the total distance D and we want the x component, then Dx will equal to D cos
theta (Dx=D cos O) and Dy is equal to D sin theta (Dy=D Sin O). If you're not comfortable
with the related Trigonometry, refer back to my MCAT Math videos which you can find
on my website at http://leah4sci.com/MCATMATH along with the Math cheat sheet that shows
you the shortcut for what numbers to memorize for different angles.
In this video series, I also show you that Cos we spell with an X (Cosx) and Sin (Syn)
we spell with a Y and here you can see, Cos is adjacent because it's right next to the
angle and Syn is opposite because it's the farthest away from the angle. So let's see
how to solve this without a calculator. We'll start by solving for Dx which we set
as equal to D Cosx theta (Dx = D cosx O) and I'm writing the X in there because I want
you to get use to recognizing Cos is x for these types of problems. The angle is 30 degrees
so we want Cosx of 30. Cosx of 30 is .866 but you should recognize that on the MCAT,
we're perfectly fine using the number point nine (.9). So we'll rewrite this equation,'
Dx is equal to D which is our 50 meters times 0.9 which is our estimated cosx 30 value.
The next step again without a calculator is quickly solving for 50 times 0.9 and there
are two quick ways that you can do this calculation. The first is to say that 50 times 9 is four
hundred and fifty (450) and that's because five times nine is forty five, we have the
zero so that makes it four fifty (450). Point nine (0.9) has the decimal moved one space
to the left and so for our answer we move the decimal one space to the left, that gives
us forty five meters (45m). With every answer you want to have the Math
solution and the logic solution. Verifying that the answer makes sense, we'll help
you catch careless mistakes if you did the calculations incorrectly. We predicted that
the x value will be something close to 50 meters, 45 is very close so we're good to
go. The second way to solve this is using my factor
of Ten trick which I teach on my MCAT Math video three and that's the idea of times
ten divided by ten or moving the decimal. If we have 50 x 0.9, we wanna get rid of the
decimal because it's confusing. We can multiply one number by ten divide the other by ten
or simply move one decimal to the right, one to the left. Right, left, new equation is
five times nine (5 x 9), this is forty five and I know I'm good to go. That was the
x value 45 meters. Now let's do the same thing for the Y value.
D of Y is equal to D Syn theta (Dy = D syn O), notice we spell that with the y again.
D is 50 meters, Syn of 30 degrees is .5 or one over two (1/2). If your number is clean,
I prefer to use a fraction because then I can say that 50 over 2 is 25 rather than try
to calculate 50 times point five (50 x.5). And a good way to recognize this is 25 times
2 is 50 or the value of 2 quarters is 50 cents. If you have half of 2 quarters that's 1
quarter, that's 25 cents. Again, we're just relating the numbers to what we know
so that we can quickly solve this without to write out the math in a long and complicated
way. And so for the y value we have 25 meters which
is a lot farther from 50 as we initially predicted. After you finish a lengthy calculation, always
go back to the question to make sure you answered it correctly. The question asked, how far
did the bird fly with respect to the ground, that is our x value, the bird flew 45 meters
with respect to the ground. How high is the tree branch, that's our Y value at 25 meters.
You maybe faced with the question, or you're given the x or the y component and ask to
find the result or ask to find the hypotenuse or the total distance traveled by this object.
For example, in this question we have: A catapult is launched at a 45 degrees angle
and travels a total of 4.9m with respect to the ground. Find the total distance.
The numbers we're given here are 45 degrees, we're told that it's 4.9 meters with respect
to the ground, so we know that our x value is equal to 4.9 meters. To solve for the total
distance, we now have to take the Vector concept but solve for it backwards. On your MCAT,
you don't want to waste time drawing diagrams but I'll draw on here for the sake of helping
you follow along with this equation. Forty five degrees is somewhere in the middle
of ninety so we'll say that this is our forty five degree angle. We're also told
that the x value is equal to 4.9 meters and the resultant value is unknown. If you think
back to how we solve this earlier, the x component, which we'll call Dx is equal to the resultant
or the distance times the cos of the theta value and Cos we spell with an x. We're
not solving for x here, we're solving for the Resultant.
Let's replace Cosx of 45 degrees with the value that we memorize, point seven and subtract
both sides by point seven to isolate our unknown. Anytime you're solving for an equation where
your unknown is not alone, first, isolate your unknown, rewrite your equation into something
simple and doable then start plugging your numbers because you'll save a lot of time
and confusion this way. And here's the equation we're solving
for, the Resultant is equal to the distance on the x direction which is 4.9 meters divided
by 0.7 which is the cos of 45 value and then we solve. But how do you solve this in the
MCAT without a calculator? This is where tricks are really really important.
The first thing I need to do is get rid of my decimals because 4.9 divided by 0.7, in
my head it just doesn't work. There are too many decimals and it's too confusing.
We'll use the trick of multiplying the top and bottom by ten or simply moving the decimals
to the right. If I move the decimals one space for both the numerator and the denominator,
I haven't changed the overall value of this number. If you don't believe me, punch them
both into a calculator and then come back. The number we're solving for now is 49 divided
by 7. The next thing we want to do is see if we
recognize this. 49 divides perfectly by 7 because 7 times 7 is 49. So if we take 7 out
of 49 we get 7. We take 7 out of 7 we get 1. Seven over one is simply equal to 7 and
that's our Resultant, seven meters (7m). Be sure to join me on the next video where
I show you how to add Vectors and Vector components, MCAT style.
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