Scene #

Description

Narration

1

A problem is showing on the screen. The narrator reads the problem out loud.

You’ll often be asked to add two vectors that are in different directions. In order to do that, you need to break each of those vectors up into their component vectors. Let’s look at a problem that requires us to do that.

A helicopter travels 2.47 kilometers at an angle of 37° above horizontal. The helicopter then changes direction, and travels 7.12 kilometers at an angle of 21° above horizontal. What is the magnitude and direction of the helicopter’s total displacement?

2

Underneath the problem the narrator draws an arrow that faces up at a diagonal. He writes 2.47 km along the arrow. At the bottom of the arrow he draws a dotted line to represent the angle measurement. He writes 37 degrees in the corner of the angle. He draws another arrow extending from the top of his first arrow and labels 7.12 km. At the base of his second arrow he draws another dotted line going horizontal and labels the corner made as 21 degrees.

Let’s begin by sketching this problem out so we can better understand it. Initially, the helicopter travels 2.47 kilometers at an angle of 37° above horizontal, so let’s draw that like that. We’ll label it 2.47 kilometers, and we’ll show that the angle is 37 degrees. Next, the helicopter changes direction, and travels 7.12 kilometers, which we’ll draw like that, at an angle of 21 degrees above horizontal. So we can see that in order to add these vectors together, we’re going to need to break them up into their x- and y-components, so let’s do that separately for each vector.

3

The narrator focuses on the first arrow he drew. He extends the x line in red and labels it d-sub-x. In blue he adds the y line to form a triangle and labels it d-sub-y. He then uses cosine ration to solve for the x direction writing down his steps as he says them.

He writes down the steps to use the sine ratio to solve for y.

Let’s begin with the 2.47 kilometer vector. The x-component is going to look like this, and we’ll call that displacement-sub-x. The y-component is going to look like that, and that will be the displacement in the y-direction.

So let’s solve for each of those. To solve for the displacement in the x-direction, let’s use the cosine ratio, which says cosine of 37 degrees equals the adjacent side, which is displacement in the x-direction, divided by the hypotenuse, which is 2.47 kilometers. Multiplying both sides by 2.47 kilometers, you get d-sub-x equals 1.97 kilometers.

Now let’s solve for the displacement in the y-direction. For that, we’re going to use the sine ratio. So sine of 37 degrees equals the opposite side, d-sub-y, divided by the hypotenuse, which is 2.47 kilometers. Multiplying both sides by 2.47 kilometers, we get the displacement in the y-direction is equal to 1.49 kilometers.

4

The narrator now separates the second arrow and extends the x line out in red adding the y line moving up in blue to create a triangle. Again he uses the cosine ratio to solve for x and the sine ratio to solve for y writing down his steps as he solves them.

Now let’s go through the same process for the second leg of the helicopter flight, which is 7.12 kilometers at an angle of 21 degrees. The x-component of this displacement is going to look like this, we’ll call that d-sub-x. The y-component of this displacement will look like that, and again we’ll call that d-sub-y. Now let’s go through the same process for this vector.

Again, we’re going to use the cosine ratio to solve for the x-displacement. This time it’s going to be cosine of 21 degrees equals d-sub-x divided by the hypotenuse, which is 7.12 kilometers. Multiplying both sides by 7.12 kilometers, we get d-sub-x for this leg is equal to 6.65 kilometers.

To solve for the displacement in the y-direction, we’re again going to use the sine ratio, which says sine of 21 degrees is equal to the displacement in the y-direction, d-sub-y, over the hypotenuse, which is 7.12 kilometers. Multiplying both sides by 7.12 kilometers, we get the displacement in the y-direction is equal to 2.55 kilometers.

5

The narrator adds the two x displacements and then adds the two y displacements.

Alright, to find our final resultant, the sum of these two vectors, we’re going to add the x- and y-components separately.

First, let’s find the sum of the x-components. Our total displacement in the x-direction is going to be equal to our first displacement in the x-direction, 1.97 kilometers, plus our second displacement in the x-direction, which was 6.65 kilometers. That gives us a displacement in the x-direction of 8.62 kilometers.

Our displacement in the y-direction, d-sub-y, is going to be equal to our first displacement in the y-direction, 1.49 kilometers, plus our second displacement in the y-direction, which was 2.55 kilometers, giving us that our displacement in the y-direction is 4.04 kilometers.

6

The narrator draws a red x arrow and puts the final displacement underneath it. He draws a blue y arrow and puts the final displacement beside it. He then finishes the triangle with a green arrow and labels it d.

Now let’s take this information to find our final resultant displacement vector. Let’s draw what we’ve learned so far. So at this point we know the final displacement in the x-direction is 8.62 kilometers, the final displacement in the y-direction is 4.04 kilometers, so we want to figure out what this resultant displacement vector is, that we’ll just call d.

7

Below the triangle the narrator writes and says the steps to solve for the Pythagorean Theorem. Finally he walks through how to solve for theta using the tangent ratio.

Let’s begin by solving for the magnitude of that vector, and to do that we can use the Pythagorean theorem. We’re going to say the magnitude of the displacement vector is equal to the square root of 8.62 squared plus 4.04 squared. And that gives us the magnitude of our displacement vector is equal to 9.52 kilometers.

Now let’s solve for the angle, theta. To do that, we’re going to use the tangent ratio, which says tangent of theta is equal to opposite over adjacent. So we can say tangent of theta is equal to the opposite side, 4.04, over the adjacent side, which is 8.62. We can take the inverse tangent of both sides, enter that into your calculator, and you’ll see that theta equals approximately 25 degrees.

So, our final displacement vector is equal to 9.52 kilometers, at an angle of 25 degrees above horizontal.