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# What are velocity components?

Learn how to simplify vectors by breaking them into parts.

## Why do we break up vectors into components?

Two-dimensional motion is more complex than one-dimensional motion since the velocities can point in diagonal directions. For example, a baseball could be moving both horizontally and vertically at the same time with a diagonal velocity $v$. We break up the velocity vector, $v$, of the baseball into two separate horizontal, ${v}_{x}$, and vertical, ${v}_{y}$, directions to simplify our calculations.
Trying to tackle both the horizontal and vertical directions of a baseball in one single equation is difficult; it’s better to take a divide-and-conquer approach.
Breaking up the diagonal velocity $v$ into horizontal ${v}_{x}$ and vertical ${v}_{y}$ components allows us to deal with each direction separately. Essentially, we'll be able to turn one difficult two-dimensional problem into two easier one-dimensional problems. This trick of breaking up vectors into components works even when the vector is something other than velocity, for example, forces, momentum, or electric fields. In fact, you'll use this trick over and over in physics, so it's important to get really good at dealing with vector components as soon as possible.

## How do we break a vector into components?

Before we talk about breaking up vectors, we should note that trigonometry already gives us the ability to relate the side lengths of a right triangle—hypotenuse, opposite, adjacent—and one of the angles, $\theta$, as seen below.
$\mathrm{sin}\phantom{\rule{0.167em}{0ex}}\theta =\frac{\text{opposite}}{\text{hypotenuse}}$
$\mathrm{cos}\phantom{\rule{0.167em}{0ex}}\theta =\frac{\text{adjacent}}{\text{hypotenuse}}$
$\mathrm{tan}\phantom{\rule{0.167em}{0ex}}\theta =\frac{\text{opposite}}{\text{adjacent}}$

When we break any diagonal vector into two perpendicular components, the total vector and its components—$v,{v}_{y},{v}_{x}$—form a right triangle. Because of this, we can apply the same trigonometric rules to a velocity vector magnitude and its components, as seen below. Notice that ${v}_{x}$ is treated as the adjacent side, ${v}_{y}$ as the opposite, and $v$ as the hypotenuse.
$\mathrm{sin}\phantom{\rule{0.167em}{0ex}}\theta =\frac{{v}_{y}}{v}$
$\mathrm{cos}\phantom{\rule{0.167em}{0ex}}\theta =\frac{{v}_{x}}{v}$
$\mathrm{tan}\phantom{\rule{0.167em}{0ex}}\theta =\frac{{v}_{y}}{{v}_{x}}$

Note that the $v$s in these formulas refer to the magnitudes of the total velocity vector, the total speed, and can therefore never be negative. The individual components ${v}_{x}$ and ${v}_{y}$ can be negative if they point in a negative direction. The convention is that left is negative for the horizontal direction, $x$, and down is negative for the vertical direction, $y$.

## How do you determine the magnitude and angle of the total vector?

We saw in the previous sections how a vector magnitude and angle can be broken up into vertical and horizontal components. But what if you start with some given velocity components: ${v}_{y}$ and ${v}_{x}$? How could you use the components to find the magnitude $v$ and angle $\theta$ of the total velocity vector?
Finding the magnitude of the total velocity vector isn't too hard since for any right triangle the side lengths and hypotenuse will be related by the Pythagorean theorem.
${v}^{2}={{v}_{x}}^{2}+{{v}_{y}}^{2}$
By taking a square root, we get the magnitude of the total velocity vector in terms of the components.
$v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}$
Also, if we know both components of the total vector, we can find the angle of the total vector using $\text{tan}\theta$.
$\mathrm{tan}\phantom{\rule{0.167em}{0ex}}\theta =\frac{{v}_{y}}{{v}_{x}}$
By taking inverse tangent, we get the angle of the total velocity vector in terms of the components.
$\theta ={\mathrm{tan}}^{-1}\left(\frac{{v}_{y}}{{v}_{x}}\right)$

## What's confusing about vector components?

When using $\theta ={\mathrm{tan}}^{-1}\left(\frac{{v}_{y}}{{v}_{x}}\right)$, the fact that we put ${v}_{y}$ on top as the opposite side and the ${v}_{x}$ on the bottom as the adjacent side means that we are measuring the angle from the horizontal axis. It seems like figuring out how to draw the angle could be confusing, but here are two good tips:
Assuming we have selected right/up as the positive directions, if the horizontal component ${v}_{x}$ is positive, the vector points rightward. If the horizontal component, ${v}_{x}$ is negative, the vector points leftward.
Again, asumming we have selected right/up as the positive directions, if the vertical component ${v}_{y}$ is positive, the vector points upward. If the vertical component ${v}_{y}$ is negative, the vector points downward.
So, for example, if the components of a vector are and , the vector must point leftward—because ${v}_{x}$ is negative—and up—because ${v}_{y}$ is positive.
Concept check: If a paper airplane has the velocity components and , which direction is the paper airplane moving—assuming we choose right and up as positive directions?

## What do solved examples involving vector components look like?

### Example 1: Bend it like Beckham

A soccer ball is kicked up and to the right at an angle of 30${}^{\circ }$ with a speed of 24.3 m/s as seen below.
What is the vertical component of the velocity at the moment shown?
What is the horizontal component of the velocity at the moment shown?
To find the vertical component of the velocity, we'll use $sin\theta =\frac{\text{opposite}}{\text{hypotenuse}}=\frac{{v}_{y}}{v}$. The hypotenuse is the magnitude of the velocity 24.3 m/s, $v$, and the opposite side to the angle 30${}^{\circ }$ is ${v}_{y}$.
$\mathrm{sin}\theta =\frac{{v}_{y}}{v}\phantom{\rule{2em}{0ex}}\text{(Use the definition of sine.)}$
${v}_{y}=v\mathrm{sin}\theta \phantom{\rule{2em}{0ex}}\text{(Solve for vertical component.)}$
To find the horizontal component, we'll use $\mathrm{cos}\theta =\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{{v}_{x}}{v}$.
$\mathrm{cos}\theta =\frac{{v}_{x}}{v}\phantom{\rule{2em}{0ex}}\text{(Use the definition of cosine.)}$
${v}_{x}=v\mathrm{cos}\theta \phantom{\rule{2em}{0ex}}\text{(Solve for horizontal component.)}$

### Example 2: Angry seagull

An angry seagull is flying over Seattle with a horizontal component of velocity and a vertical component of velocity .
What is the magnitude of the total velocity of the seagull?
What is the angle of the total velocity?
Assume right/up are positive, and assume all angles will be measured counterclockwise from the positive x axis.
We'll use the Pythagorean theorem to find the magnitude of the total velocity vector.
${v}^{2}={v}_{x}^{2}+{v}_{y}^{2}\phantom{\rule{2em}{0ex}}\text{(The Pythagorean theorem.)}$
$v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}\phantom{\rule{2em}{0ex}}\text{(Take square root of both sides.)}$
To find the angle, we'll use the definition of $\text{tangent}$, but since we now know $v$, we could have used $\text{sine}$ or $\text{cosine}$.
$\mathrm{tan}\theta =\frac{{v}_{y}}{{v}_{x}}\phantom{\rule{2em}{0ex}}\text{(Use the definition of tangent.)}$
$\theta ={\mathrm{tan}}^{-1}\left(\frac{{v}_{y}}{{v}_{x}}\right)\phantom{\rule{2em}{0ex}}\text{(Inverse tangent of both sides.)}$
$\theta ={30.6}^{\circ }\phantom{\rule{2em}{0ex}}\text{(Calculate and celebrate!)}$
Since the vertical component is , we know the vector is directed down, and since , we know the vector is directed right. So, we will draw the vector in the fourth quadrant.
The seagull is moving at an angle of ${30.6}^{\circ }$ below the horizontal.

## Want to join the conversation?

• Although the same question has been asked, i didn't quite get your conclusion. When finding the angle of the vector, you didn't keep the y-component negative. i did and my angle was negative. Does it matter? My answer's still correct?
• In that part of the Angry Seagull question he isn't taking the exact velocities, he is taking the magnitudes of the velocities (just the number, not the sign). Magnitudes are always positive. The angle he gets is positive, because it's the magnitude of that angle. If you were to look at it on a polar graph, you could indeed list that as a negative angle, and it would indicate that the bird is flying right (positive) and down (negative).

To sum it all up, if you want to succinctly tell what direction the bird is heading, the negative angle is the correct answer [on a polar graph this would read (17.0 , -30.6 degrees)], but if all you care about is the size of the angle, the answer in the article is correct.
• What is the meaning of terminal velocity?
• How do we calculate the air resistance and in turn terminal velocity?
• But how would we know if our triangle is right angled? would we have to make a scale diagram first or?? im so confused
• Assuming you draw a horizontal and a vertical vector, as they showed, you will always end up with a right triangle. This is because the horizontal and vertical lines are perpendicular to each other, creating a right angle. Of course, you can instead draw the vectors as non-horizontal and vertical lines (as they also showed), but I think it would make the triangle much harder to use. Just remember that as long as your triangle has two lines perpendicular to each other, it is a right triangle with a right angle.
Hope this helps! Let me know if you still don't understand =)
• I notice in the presentations, you use double lines on each side of the components. What do the double lines mean?
• Hello Bob,

When we work with vectors double lines such as this ||X|| or single lines such as |X| are referring to the length of vector X.

Regards,

APD
• so this is kind of like finding the degrees of a point on the unit circle using:
sin( )
cos( )
tan( )
csc( )
sec( )
cot( )
arcsin( )
arccos( )
arctan( )
arccsc( )
arcsec( )
arccot( )

Right?
Or no?
• Sorta? But you will not need to know the list after tangent from what I have read up on the MCAT.
• Can I have more than two vectors that affects my velocity?
• Yes. There is a 3rd axis of movement known as z used in 3-D diagrams. Movement on this axis implies going in or out of the page. Also you can have an infinite quantity of vectors affecting velocity as long as you account for each individual vector's velocity.
• Can any one give me the all formulas of Projectile motion chapter
(1 vote)
• Is arcsin the same as sin^-1 ? And for the rest too?