# Velocity vectors and trigonometry

I am trying to learn about velocity vectors but this word problem is confusing me.

A boat is going 20 mph north east, the velocity u of the boat is the durection of the boats motion, and length is 20, the boat's speed. If the positive y axis represents north and x is east the boats direction makes an angle of 45 degrees. You can compute the components of u by using trig

$$u_1 = 20 \cos 45$$ $$u_2 = 20 \sin 45$$

Why? How did this happen? Why sin and why cos? What does this represent? Why two points? What are these two points? It says that these are $R^2$ which I am not sure what that means and my book does not explain. I think t he R means all real numbers and the squared is referencing 2d maybe so x and y but the book doesn't say so I am not so sure. My book mentions none of these things.

-
$\mathbf{R}^2 = \mathbb{R}^2$ is 2-D space. (More specifically, it's the set of all ordered pairs $(x, y)$ such that $x$ and $y$ are real. –  apnorton Sep 4 '13 at 0:06
Doesn't hollow R mean all real numbers? So all real numbers squared? –  Paul the Pirate Sep 4 '13 at 0:14
The "hollow R" ($\mathbb{R}$) is the set of all real numbers. You can't really square a set, so the notation $\mathbb{R}^2$ is shorthand for the Cartesian product $\mathbb{R}\times\mathbb{R}$, which is an operation on sets, not on the numbers in the set. For more information on Cartesian products, see here. –  apnorton Sep 4 '13 at 0:25

$$u_1=20cos(45)$$ is the component of the velocity vector in the x-direction. $$u_2=20sin(45)$$ is the component of the velocity vector in the y-direction.

If you add these two vectors head to tail, then what you get is the velocity vector of the boat (25mph, north-east, i.e. at a 45 degree angle to both the x and y-axes).

Draw the vector out on the x-y plane starting at the origin, now draw the x-component of it and y-component of it. Now you know your hypotenuse is the velocity of the boat in the direction it's travelling in and the angle it makes with x-axis. You use basic trigonometry to obtain the values above (sine = opposite/hypotenuse, cos = adjacent/hypotenuse).

$R^2$ refers to the all points in the xy-plane (I think this is a good enough definition for this purpose). $u_1$ and $u_2$ are vectors.

-
I don't follow the basic trig suggestion, I only have 1 line and 2 angles. –  Paul the Pirate Sep 4 '13 at 0:16
You have a line from the vector. Now, draw a pair of $x$- and $y$-axes at the tail of the vector. The angle from the $x$-axis to the vector is $45^\circ$. –  apnorton Sep 4 '13 at 0:26
I don't follow, why do I have a line coming out of the vector? –  Paul the Pirate Sep 4 '13 at 15:13

. Dear Mr. The Pirate,

(I made a worksheet to answer your question. Click "Go to student worksheet". The arrow keys at the bottom walk through the steps.)

Sorry I left you oarless.

-- Mr. Land

P.S. When I refer to $\mathbb{R}^2$, I say, "(coordinates in) (the x-y) plane". Good luck and don't let the maths get you down.

-

First note that that $u_1$ and $u_2$ aren't points, actually the question explains, $u_1$ and $u_2$ are components of the velocity.

Velocity is a vector and using these method we split it into 2 other vectors,actually we make projection on the x-axis and the y-axis. The one is the x-velocity, velocity in x direction and its direction is parallel to the x-axis, while the other is the y-velocity, and its direction is parallel to the y-axis. And if we add up these 2 vectors we we'll get the inital velocity vector. This split is to simplify things and to make things easier if we make some not so simple caluclations.

To make things even more easier if the speed of the boat is $20 \text{ m/s}$ and the angle is $45^{\circ}$ the x-velocity and the y-velocity will be same, which means that the vectors will have same magnitude of $10\sqrt{2}$. This means that after time of 1 second the boat will change its x-position for $10\sqrt{2}\text{ m}$ and also will change its y-position for $10\sqrt{2}\text{ m}$ and will travel $20\text{ m}$.

If you have a little knowledge in trigonometry, you'll certainly know about the unit circle. If we take a random angle in the unit circle, the the cosine function of that angle will give us the projection on the x-axis or the x-velocity in this case. While the sine function we'll give us the projection on y-axis.

$\mathbb{R}^n$ represents all ordered pairs $(x_1,x_2,...,x_n)$, such as $x_1,x_2,...,x_n \in \mathbb{R}$. While $n$ represent the dimension we work. If $n=2$ means that we work in $2D$, which is obvious for this case, because we could think of the water surface the boat floats in as a plane. And the ordered pair $(x,y)$ will represent the position of the boat.

Let's leave the vector and just look at the triangle in this picture:

This is a right triangle and the $\angle BCA = 50^{\circ}$. Assume that we know the lenght of the hypotenuse and it's $10\text{ cm}$, we can easily find the length of the other sides.

If we want to find the length of the opposite side we just calculate:

$$20\text{ cm} \times \sin{50^{\circ}}$$

While if we want to find the length of the adjacent side realtive to teh angle we calculate:

$$20\text{ cm} \times \cos{50^{\circ}}$$

Now think like this. The point is at the point $C$ of the triangle and it's heading to the point $A$. We could think of the side of the triangle as a vector and the magnitude of the vector will be the length of the hypotenuse.

If we want to find the projection on the y-axis or the velocity in y-direction we need to find the magnitude of the vector $BA$, i.e. the length of the side $AB$, which is opposite side of the angle. We showed how to find using the upper equation so here's once again:

$$\text{Length of AB} = \text{Length of hypothenuse} \times \text{Sine of the opposite angle}$$

In other words:

$$\text{Velocity in y-direction} = \text{Velocity} \times \text{Sine of the opposite angle}$$

If we want to find the projection on the x-axis, i.e x-velocity we need to find the length of the side $BC$. Because this side is adjacent ot the given angle we need to use the cosine function, so we calculate:

$$\text{Length of BC} = \text{Length of hypothenuse} \times \text{Cosine of the adjacent angle}$$

In other words:

$$\text{Velocity in x-direction} = \text{Velocity} \times \text{Cosine of the adjacent angle}$$

-
I have forgotten that trig trick, what is it called? –  Paul the Pirate Sep 4 '13 at 2:50
I think it's called vector decomposition. –  Stefan4024 Sep 4 '13 at 9:39
No the part where you said If you have a little knowledge in trigonometry, you'll certainly know about the unit circle. If we take a random angle in the unit circle, the the cosine function of that angle will give us the projection on the x-axis or the x-velocity in this case. While the sine function we'll give us the projection on y-axis. –  Paul the Pirate Sep 4 '13 at 15:14
In other words, where did you magic up the term $10\sqrt2$ –  Paul the Pirate Sep 4 '13 at 15:18
$$\sin{45^{\circ}} = \frac{\sqrt{2}}{2}$$. Also we have $$\cos{45^{\circ}} = \frac{\sqrt{2}}{2}$$ So $20 \times \sin{45^{\circ}} = \frac{20 \times \sqrt{2}}{2} = 10\sqrt{2}$ Simularly for the cosine function –  Stefan4024 Sep 4 '13 at 16:38