Good morning. Bobby, please break the initial velocity vector of 35.0 meters per second at 25.0
degrees above the horizontal into its components. ♪ Flipping Physics ♪
Okay. We just use soh cah toa to do that. Sine theta equals opposite over hypotenuse
or velocity initial in the y direction over velocity initial. That means velocity initial
in the y direction equals velocity initial times sine theta or 35 sine 25 or 14.7916
or 14.8 meters per second with 3 sig figs. Cosine theta equals adjacent over hypotenuse or velocity
initial in the x direction over velocity initial. That means velocity initial in the x direction
equals velocity initial times cosine theta or 35 cosine 25 or 31.7208 or 31.7 meters per second.
Thank you Bobby. Those are the components of the initial velocity vector. But class,
what does it mean that the initial velocity in the y-direction of 14.8 meters per second
and the initial velocity in the x direction of 31.7 meters per second are the components of the
initial velocity of 35.0 meters per second at an angle of 25.0 degrees above the horizontal?
Those are the components of the original vector.
(Right. But what does that mean?) The components are the
pieces of the original vector. When you add the components together,
you get the original vector. So the velocity initial in
the x-direction plus the velocity initial in the y-direction equals the velocity initial?
Yes! Yeah. Yup. Really? 14.8 plus 31.7 equals 35?
Yes? Uh…
No, you left off the direction. Right. 14.8 meters per second in
the x-direction plus 31.7 meters per second in the y-direction equals 35.0 meters per
second at 25.0 degrees above the horizontal. But what is wrong with velocity initial in
the x-direction plus velocity initial in the y-direction equals velocity initial?
He left the vector symbol off of each velocity. When you add the vector symbols to that equation,
the equation is correct. Oh!!
Absolutely correct. Now, using the Pythagorean Theorem and tangent theta, you can prove that the
components of the vector add up to the original vector, however, I am not going to talk my way
through all of that right now. Instead, I am going to express the initial velocity vector
using its components and unit vectors. In other words, this velocity initial equals 31.7 i,
with a unit vector hat, plus 14.8 j, with a unit vector hat, all in meters per second.
i, with a unit vector hat, j, with a unit vector hat, and k, with a unit vector hat, are vectors
with a magnitude of 1 unit in the x, y, and z directions respectively. That is why i, j, and k
are called “unit” vectors. To be absolutely clear, the following two expressions for initial velocity
identify the same vector, one uses magnitude and direction, the other uses unit vectors.
(Mr.p?) Yes Bobby?
Why? (Why what?)
Why use unit vectors? I mean, why do we have two different ways to describe the same vector?
Ah, good question. To answer that, let’s do a basic vector addition example. Bo, please
read the problem, and Billy, please solve it. Determine the resultant vector when adding A,
B, and C together. A equals 2.0 i plus 2.0 j meters. B equals 1.0 i minus 3.0 j
meters. C equals 4.0 i plus 4.0 j meters. The resultant vector equals A plus B plus C.
Hold up, where are the vector symbols for A, B, and C in the first sentence of the problem?
Shouldn’t those all have vector symbols on them? Oh, sometimes vectors are written boldface
instead of with a vector symbol over the top. I bet mr.p is just reviewing that for us here.
Yep. Billy? Okay. R equals A plus B plus C.
We should start by drawing out the vector diagram. Then, substitute in the values of each vector.
And combine i's and combine j’s to get 2 plus 1 plus 4 i plus 2 minus 3 plus 4 j.
That equals 7.0 i plus 3.0 j meters. Wait, that’s it? That’s all we have to do?
Okay, that is why we will sometimes use unit vectors to describe vectors. That was really
easy. Imagine if each vector were instead given with magnitude and direction, then we would need
to break each vector into its components, and then add all the x-values together, and then add
all the y-values together, and then determine the magnitude and direction of the resultant vector.
That would take a lot longer to do than what Billy just did with unit vectors. Got it.
Bobby, that is correct. Okay, let’s do one more example. Bobby, please read and Bo, please solve.
What is the distance between the following two positions. R 1
I guess that is position 1 equals 15.0 meters at 55.0 degrees East of North.
And position 2 equals 25.0 meters at 45.0 degrees North of East.
Well, start by drawing both vectors. Okay, we need to find the difference between the two vectors,
which is delta r. And add delta r to the vector drawing. Once we have delta r,
we can determine the magnitude of delta r to get the distance between those two positions.
r 1 plus delta r equals r 2. That means delta r equals r 2 minus r 1. Now we need
to determine the components of r 1 and r 2. Actually, Bo, please express position 1
and position 2 using unit vectors and use unit vectors to solve this problem.
If he wants us to use unit vectors, then why did he give us vectors with magnitude and direction?
He’s probably just forcing us to review. He’s always doing that.
Yeah, that he is. Okay. Sine theta equals opposite
over hypotenuse Actually, we’ve done this a lot with SOH CAH TOA. Can we just hop forward
to the r cosine theta and r sine theta bit? Bo. You are correct. We have used soh cah
toa a lot and, this is going to come up a lot in our calculus-based physics class. So yes,
in our calculus-based physics class, feel free to start at the r sine or cosine theta part.
Woah. Sweet. Alright, cosine always goes with x and
sine always goes with y so position 1 equals r 1 cosine 55 i plus r 1 sine 55 j meters. That equals
8.60365 i plus 12.2872 j meters. Position 2 equals r 2 cosine 45 i plus r 2 sine 45 j meters.
Which equals 17.6777 i plus 17.6777 j meters. And now we solve for delta r. Uh.
Wait a second Bo. What?
I think you got your sine and cosine switched for position 1.
What? Yeah, the first position vector angle is
55 degrees East of North not North of East. That means sine goes with x and cosine goes with y.
Oh, right. Oops. Yeah. Please be careful to notice if your angle
is with the horizontal or the vertical. It really makes a difference. Because it is not correct that
cosine always goes with x and sine always goes with y. Again, it is not correct that cosine
always goes with x and sine always goes with y. Yeah. Well, now we can solve it. Delta r
equals position 2 minus position 1. Substitute in the unit vectors.
Combine the i values and combine the j values and we get 5.39040 i plus 9.07405 j meters for
delta r. And now we need to find the magnitude … Wait a second. Delta r is change in position.
If an object goes from position 1 to position 2, they will have gone through a displacement
of 5.39 i plus 9.07 j meters. That would be the object’s displacement!
That is correct Billy, thanks. Bo, please finish the problem.
Alright. Using the Pythagorean Theorem, we know the magnitude of delta r
equals the square root of 5.39040 squared plus 9.07405 squared which equals 10.5544
or 10.6 meters. The distance between position 1 and position 2, with 3 sig figs, is 10.6 meters.
Correct. Now, I want to talk more about this r position vector. The r position
vector is used to identify the location of an object in multiple dimensions. For example, r
position vector equals x times unit vector i plus y times unit vector j plus z times unit vector k.
This position vector identifies the location of an object in 3-dimensional space. As Billy pointed
out, we can also have the change in r position of an object, which is the displacement of the object
in 3-dimensional space. In other words, we are no longer only identifying the
position and displacement of an object in the x, y, or z directions independently,
but rather, all three directions at the same time using the r position vector.
Okay class, what then is change in r position over delta t equal to?
Velocity. Average velocity. Remember,
that is average velocity. Yeah, average velocity.
And what then is the derivative of r position with respect to time equal to?
Instantaneous velocity. Now, I know this does not look all
that different from what we were doing before, however, realize we can now identify position,
displacement, velocity, and acceleration, in 3 dimensions, all at once, using
unit vectors. The ramifications of that are pretty cool. We’ll learn more about that
later. Thank you very much for learning with me today, I enjoyed learning with you.