Leah here from leah4sci.com/MCATPHYSICS and
in this video we'll continue our discussion
on Adding and Subtracting vectors focusing
on Vectors that are given at some angle feda
where you have to find the X and Y components.
If you missed the first parts of this video,
go find it on my website at http://leah4sci.com/MCATPHYSICS.
Let's start with the simple question to
review and then go into angles.
Say you're asked to find the total displacement
if a biker travels 8km east then turns and
bikes 6km North.
Since the biker is not traveling in one direction
you can't simply add the numbers. Instead
you have to take each league of the journey
from the X and Y components and then use the
Pythagorean Theorem to find the total displacement
or the Resultant.
We'll draw a quick sketch. We have a displacement
of 8 kilometers east or to the right, followed
by 6 kilometers North. Notice that I have
the head to tail method going here. We start
at the tail and end at the head, in return
we have a brand new tail and end at the second
head.
The total Resultant Vector is a line going
from the very first tail to the very last
head and we call that R for our Resultant.
The reason I chose this example is because
on the MCAT, in addition to knowing Math equations
and how to solve them, you also want to look
out for tricks and patterns that will save
you time.
There is a triangle that has the angles 36
to 90 degrees will fall into a pattern of
what we call a 3 4 5 triangle. And the 3 4
5 triangles as follows: If you have your short
side at a length of 3, your longer side a
length of 4, the hypotenuse will be a length
of 5. If you can recognize the triangle in
your MCAT that has some factor of these numbers,
you can save yourself the trouble, save yourself
the math and get the answer right there.
Notice that we have 6 and 8. Six is two times
three; eight is two times four so we expect
the Resultant to be two times five for a total
of ten kilometers. Now let's prove it mathematically.
We'll use the components B1 for biking the
first league of the journey and B2 for biking
the second league of the journey. The first
league of the journey is 8 kilometers in the
x direction so we have B1x is equal to eight
(B1x=8). Since B1 is only in the X direction,
B1y is equal to zero.
The second league of the journey is only in
the Y direction so B2x is equal to zero, B2y
is equal to six. And so x or the sum of x
is equal to eight plus zero which is eight.
The sum of Y is equal to six plus zero which
is equal to six. And the resultant comes from
the formula X squared plus Y squared is equals
to Z squared (X^2 + Y^2 = Z^2) or Resultant
squared. You should
Know this formula and be confident enough
that you don't have to write it out this way,
instead you should simply write the Resultant
is equal to the square root of X squared plus
Y squared. In this case x squared is eight
squared or sixty four, Y squared is six squared
or thirty six. If we have sixty four and thirty
six, sixty and thirty is ninety, four and
six is ten which gives us ninety plus ten
or a hundred, that means the resultant is
the square root of one hundred which is equal
to ten and that's the number we initially
predicted.
If you're not comfortable with the multiplication
that we did here, download my Math Study Guide
Cheat Sheet which you can find on my website
at http://leah4sci.com/MCATMATH.
Here's another question that you may find
in your MCAT practice material. Now keep in
mind the MCAT will not ask you something this
tedious and detailed when it comes to Math
because they're testing you in your ability
to apply equations and concepts, they're
not looking for you to waste an hour on a
single Math calculation. The reason I'm
including this is because I want to make sure
that you're comfortable with every single
component that shows up within this question
so that you'll know what to do if you're
face something along this line. The question
says:
Find the total displacement if a dog runs
59 meters at 45 degrees to the horizontal,
then turns North running 31 meters at 30 degrees
with respect to the vertical.
This is a little bit of a tricky question
so make sure that you break it down and understand
what's going on here. For something like
this, it helps if you draw a quick sketch
and then work of the sketch rather than trying
to pick the numbers out of the story.
So we have two vectors here, the dog is running
59 meters at 45 degrees to the horizontal.
So let's start with the horizontal and then
set the angle at 45 degrees with the vector
that is 59 meters long. The second vector
is in the Northern direction for 31 meters
at a 30-degree angle with respect to the vertical.
This trips off a lot of the students so be
careful. If we draw a vertical or a Y component,
it helps if you can also see an x component
for that new set. It's almost like we're
putting two different graphs together. If
we have 30 degrees with respect to the vertical,
that's 30 degrees from the Y access and
not the X axis. If we draw a 30 degree angle
and then start to calculate, we're looking
for everything to fit into a single coordinate
system and that means we want all angles to
be with respect to a graph that comes out
of here. These are our angles and that means
that 30 degrees won't work, we need to find
the angle on the other side of that 30 degrees
and that's why I drew the X and Y components.
The angle between X and Y is 90 degrees and
so if we have 30 degrees at the top, we do
90 minus 30 and we're actually looking at
the 60 degree angle at our calculations rather
than the 30 degree angle.
Since the dog is running, we'll call our
first Vector R1 for run 1, we'll call the
second vector R2 for run 2 and we have to
now break down the individual X and Y components.
For R1, we have 59 meters at a 45-degree angle.
That means R1x is equal to 59 cos 45 and R1y
is equal to 59 sin of 45.
The reason I need to use Cos for X and Sin
for Y is because I'm referring to my trick
of spelling cos with an x and syn with a y.
I talked about this trick in my Trigonometry
video which you can find on my website at
http://leah4sci.com/MCATMATH. In that video
I also talked about the shortcut values to
memorize for the Sin and Cos values. 45 degrees
is going to be the same for Sin and Cos and
we're using the value of 0.7 and so we have
59 times 0.7 for the X and y values. Calculating
this exactly will take too long so we'll
take the shortcut of rounding 59 to 60, that's
very close and multiplying 60 times 0.7. I'll
use the trick of times ten divided by ten
or move one decimal to the right and one decimal
to the left. Also a topic covered on my Math
videos and that means I move this one to the
right and this one to the left giving me six
times seven which is forty two.
Now moving on to R2. For R2 we have 31 meters
at a 60-degree angle and that means R2x is
31 cos with an x of 60 and R2y is equal to
31 syn 60. The value to memorize, cos of 60
is 0.5,
 syn of 60 is 0.866 but it's okay
if you memorize and use the value of 0.9.
So let's calculate what we have. Thirty
one times point five, that's approximately
thirty times point five, thirty over two which
gives me a rounded value of fifteen. And thirty
one times point nine, ones again we'll use
the decimal trick move that to the right,
move that to the left that gives me three
times nine which is twenty seven (3x9=27).
That's equal to X and that's equal to
Y.
Now that we have our Vector components we
wanna add them up. So we'll have the sum
of X is equal to X1 plus X2 (ex=X1+X2). The
sum of Y is equal to Y1 plus Y2( eY=Y1+Y2).
X1 is forty two (42) and X2 is fifteen (15)
but be careful about the direction. Notice
that R1 goes to the right which is a positive
direction, R2 goes to the left which is a
negative x direction and so it's really
plus negative 15 or simply minus 15. Forty
two minus fifteen is a little hard to do,
you can try forty minus fifteen which is twenty
five then add back the two giving you twenty
seven (27). Y1 is forty two Y2 is twenty seven,
forty and twenty is sixty, two and seven is
nine giving us a value of sixty nine (69).
And now we solve for the Resultant. The resultant
is the square of X squared plus Y squared,
in this case the square of twenty seven squared
plus sixty nine squared ((R=v(?27?^2+?69?^2
)) . These numbers are very hard to do the
long way so we're going to round. And since
we're rounding everything up, we're going
to assume that our answer will be slightly
less. Remember we're making our numbers
bigger so make sure you compensate by making
your answers slightly smaller. We'll square
thirty and seventy.
While we don't know 30 squared I know three
squared is nine and then I add two zeros,
one for each of the thirty and that gives
me nine hundred. I do the same for seventy.
Seven squared is forty nine and I add two
zeros, one for each seventy that I'm using.
Now we add Nine hundred and forty nine hundred.
If we look at forty nine as fifty, that's
fifty nine so the answer will be fifty eight
and then we have fifty eight hundred (5800).
And finally we take the square root of that.
I don't know the square root of fifty eight
hundred but I do know that it's equal to
the square root of fifty eight times the square
root of a hundred. The square root of fifty
eight is not a number that I know, so I try
to find a recognizable square below and above
the number that I'm given.
The square root of forty nine is less than
the square root of fifty eight which is less
than the square root of sixty four. Rad forty
nine is seven, rad sixty four is eight. So
we'll guesstimate that our answer is approximately
seven point five. The square root of one hundred
is ten, that's simple enough and our final
calculation is seven point five times ten.
Whenever you multiply by ten we move the decimal
to the right which gives us a Resultant of
seventy five meters (R=75m).
Keeping in mind that we did a lot of rounding
to get to where we are, I punched the initial
equation in to the calculator and got an answer
of Seventy three point three (73.3) which
given the amount of rounding we did is very
very close. Now remember, on the MCAT you
will not have this many steps in a single
question but you do have to know how to approach
this in case you're faced with a question
that is similar to any part of this long question
that we just solved.
Be sure to join me in the next video where
we start our discussion in Kinematics by talking
about speed and velocity as it appears on
the MCAT.
Are you stuck on a specific MCAT topic? I
offer Private Online Tutoring where I focus
on your needs to strengthen your individual
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leah4sci.com/MCATTutor.
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