- [Instructor] We're told a boat is traveling
at a speed of 26 kilometers per hour in a direction
that is a 300 degree rotation from East.
At a certain point it encounters a current
at a speed of 15 kilometers per hour in a direction
that is a 25 degree rotation from East.
Answer two questions about the boat's velocity
after it meets the current.
Alright, the first question is,
what is the boat's speed after it meets the current?
And it says, round your answer to the nearest 10th.
You can round intermediate values to the nearest 100th.
And what is the direction of the boat's velocity
after it meets the current?
And they say the same, well, they actually here it say,
round your answer to the nearest integer
and you can round intermediate values to the nearest 100th.
So like always pause this video
and see if you can work through this.
All right, now let's work on this together.
So first let's visualize each of these vectors.
We have this vector 26 kilometers per hour in a direction
that is a 300 degree rotation from East.
And we have this vector 15 kilometers per hour
in a direction that is a 25 degree rotation from East.
And so let me draw some axes here.
So let's say that is my Y-axis.
And then let's say that this over here is my X-axis.
And then that first vector 300 degree rotation from East,
East is in the positive X direction.
This would be 90 degrees, 180 degrees, 270 degrees.
I'm going counter-clockwise cause that's the convention
for a positive angle.
And then we'd go a little bit past 270,
we would go right, right over there.
And the magnitude of this vector is 26 kilometers per hour.
I'll just write a 26 right over there.
And then this other vector which is the current
15 kilometers per hour in a direction
that is a 25 degree rotation from East.
So 25 degree rotation might be something like this
and it's going to be shorter.
It's 15 kilometers per hour.
So, it's going to be roughly about that long.
I'm obviously just approximating it
and I'll just write 15 there for its magnitude.
So we can visualize what the boat's speed
and direction it is after it meets the current.
It's going to be the sum of these two vectors.
And so if we wanted to sum these two vectors
we could put the tail of one at the head of the other.
And so let's shift this blue vector down here.
So it's at the head of the red vector.
So it would be something like this.
And so our resulting speed after it meets the current
would look something like this.
We've seen this in many other videos so far
but we don't want to just figure it out visually.
We want to actually figure out its actual speed
which would be the magnitude of this vector
and its actual direction.
So what is the angle?
And we could say it as a positive angle.
So what the rotation, the positive rotation
from the positive X-axis or from due East.
So to do that, what I'm going to do is represent each
of our original vectors in terms of their components.
And so this red vector up here
and we've done this multiple times explaining the intuition.
It's X component is going to be its magnitude
26 times the cosine of this angle,
cosine of 300 degrees.
And it's Y component is going to be 26 times the sine
of 300 degrees.
If that's unfamiliar to you, I encourage you to review it
in other videos where we first introduced
the notion of components, it comes straight out
of our unit circle definition of trig functions.
And similarly, this vector right over here
it's X component is going to be its magnitude times
the cosine of 25 degrees.
And it's Y component is going to be 15 times the sine
of 25 degrees.
And now when we have it expressed this way,
if we want to have the resulting vector,
let's call the resulting vector S
for maybe the resulting speed.
Its components are going to be the sum of each of these.
So we can write it over here.
Vector S is going to be equal to
it's going to be the X component of this red vector
of our original speed vector.
So, 26 cosine of 300 degrees
plus the X component of the current.
So, 15 times cosine of 25 degrees and then the Y components.
Once again, I add the corresponding Y components
26 sine of 300 degrees
plus 15 sine of 25 degrees.
And now we could use a calculator to figure out
what these are, to say what these approximately are.
So first the X component, we're going to take
the cosine of 300 degrees, times 26,
plus I'll open a parenthesis here.
We're going to take the cosine of 25 degrees, times 15,
close our parentheses.
And that is equal to 26.59 if I round to the nearest 100th.
26.59.
And now let's do the Y component.
We have the sine of 300 degrees, times 26,
plus I'll open parentheses,
the sine of 25 degrees times 15, close parentheses,
is equal to negative 16.18 to round to the nearest 100th.
Negative 16.18.
And let's just make sure that this makes intuitive sense.
So, 26.59.
So we're going to go forward in this direction
26.59 on the X direction.
And then we go negative 16.18 in the Y direction.
So this does seem to match our intuition
when we tried to look at this visually.
So we now have the X and Y components
of the resulting vector
but that's not what they're asking for.
They're asking for the speed which would be the magnitude
of this vector right over here.
And so I could write the magnitude of that vector
which is going to be its speed.
We'll just use the Pythagorean theorem here.
It is going to be the square root of this squared
plus this squared, because once again
this forms a right triangle here.
And we review this in other videos,
it's going to be the square root of 26.59 squared
plus negative 16.18 squared which is approximately equal to,
they want us to round to the nearest 10th,
26.59 squared plus.
And it doesn't matter that there's a negative here
cause I'm squaring it.
So I'll just write 16.18 squared is equal to that.
And then we want to take the square root of that.
We get 31 point, if we round to the nearest 10th, 31.1.
So it's approximately 31.1 and we'll write the units here,
kilometers per hour is the speed the boat's speed
after it meets the current.
And now the second question is what is the direction
of the boat's velocity after it meets the current?
Well, one way to think about it is
if we look at this angle right over here
which would tell us the direction
the tangent of that angle, theta, let me write this down.
Tangent of that angle theta.
We know your tangent is your change in Y
over your change in X.
You can even view it as the slope
of this vector right over here.
We know what our changes in X or Y are.
Those are X and Y components.
So it's going to be our change in Y
which is negative 16.18, over 26.59,
our change in X.
And so to solve for theta, we could say
that theta will be equal to the inverse tangent.
And we'll have to think about this for a second
because this might not get us the exact theta that we want
because the inverse tangent function
is going to give us something between positive 90 degrees
and negative 90 degrees.
But the number we want, actually it looks
like it's going to be between 270 and 360 degrees
because we're doing a,
we want to think about a positive rotation
instead of a negative one but let's just try to evaluate it.
The inverse tan of this,
of negative 16.18, over 26.59.
16.18 negative
divided by 26.59 is equal to this.
And now I am going to take the inverse tangent of that.
And that gets us negative 31 degrees, which makes sense.
This looks intuitive sense
that if you were to do a clockwise rotation
which would be a negative angle from the positive X-axis
it looks like what we drew, but let's just go
with the convention of everything else here.
And let's try to have a positive angle.
So what we can do is add 360 degrees to that
to make a full rotation around.
And we essentially have the equivalent angle.
So let's add 360 to that to get that right over there.
So if we round to the nearest integer
we're looking at approximately 329 degrees.
So theta is approximately 329 degrees.
So here, when I said theta is equal to this I could write
theta is going to be equal to this plus 360 degrees.
Now what's interesting is, I was able to add 360 degrees
to get to the exact same place.
If we had a situation where our angle was actually
this angle right over here not the situation
that we actually dealt with,
but if it was in the second quadrant,
we would have gotten this theta.
And we would have had to be able to realize that,
hey we're dealing with the second quadrant
that has the same slope.
So instead of adding 360 degrees
we would have added 180 degrees.
And we've also covered that in other videos as well.