Vector word problem: resultant velocity | Vectors | Precalculus | Khan Academy

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. All right, the first question is: What is the boat's speed after it meets the current? It says round your answer to the nearest tenth. You can round intermediate values to the nearest hundredth. And what is the direction of the boat's velocity after it meets the current? They say the same. Well, they actually, here it says round your answer to the nearest integer, and you can round intermediate values to the nearest hundredth.

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.

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. Then that first vector, 300 degree rotation from east, is in the positive x direction. This would be 90 degrees, 180 degrees, 270 degrees. I'm going counterclockwise because that's the convention for a positive angle. And then we'd go a little bit past 270. We would go right over there, and the magnitude of this vector is 26 kilometers per hour.

I'll just write it 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 is after it meets the current. It's going to be the sum of these two vectors. If we wanted to sum these two vectors, we could put the tail of one at the head of the other. 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? We could say it is a positive angle. So what is the rotation, the positive rotation from the positive x-axis, or from due east? To do that, what I'm going to do is represent each of our original vectors in terms of their components.

So this red vector up here, and we've done this multiple times explaining the intuition, its x component is going to be its magnitude 26 times the cosine of this angle, cosine of 300 degrees. And its 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 introduce the notion of components. It comes straight out of our unit circle definition of trig functions.

Similarly, this vector right over here, its x component is going to be its magnitude times the cosine of 25 degrees, and its y component is going to be 15 times the sine of 25 degrees.

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 times the 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 up parentheses 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 hundredth, 26.59.

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, closed parentheses, is equal to negative 16.18, to round to the nearest hundredth, 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 in the x direction, and then we go negative 16.18 in the y direction. So this does seem to match our intuition when we try 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. So I could write the magnitude of that vector, which is going to be its speed. We 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.

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 tenth, 26.59 squared plus... and it doesn't matter that there's a negative here because 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 tenth, 31.1. So it's approximately 31.1. And we'll write the units here: kilometers per hour is the speed of the boat after it meets the current.

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 our 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.

To solve for theta, we could say that theta will be equal to the inverse tangent, and we have to think about this for a second because this might not get us the exact theta that we want. The inverse tangent function is going to give us something between positive 90 degrees and negative 90 degrees, but the number we want actually looks like it's going to be between 270 and 360 degrees because 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. That gets us negative 31 degrees, which makes 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. 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 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.