What are velocity 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 $v$v. We break up the velocity vector, $v$v, of the baseball into two separate horizontal, $v_x$v, start subscript, x, end subscript, and vertical, $v_y$v, start subscript, y, end subscript, 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 $v$v into horizontal $v_x$v, start subscript, x, end subscript and vertical $v_y$v, start subscript, y, end subscript 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.

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, $\theta$theta, as seen below.

When we break any diagonal vector into two perpendicular components, the total vector and its components—$v, v_y, v_x$v, comma, v, start subscript, y, end subscript, comma, v, start subscript, x, end subscript—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 $v_x$v, start subscript, x, end subscript is treated as the adjacent side, $v_y$v, start subscript, y, end subscript as the opposite, and $v$v as the hypotenuse.

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

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: $v_y$v, start subscript, y, end subscript and $v_x$v, start subscript, x, end subscript? How could you use the components to find the magnitude $v$v and angle $\theta$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.

By taking a square root, we get the magnitude of the total velocity vector in terms of the components.

Also, if we know both components of the total vector, we can find the angle of the total vector using $\text{tan}\theta$start text, t, a, n, end text, theta.

By taking inverse tangent, we get the angle of the total velocity vector in terms of the components.

When using $\theta=\tan^{-1}(\dfrac{{v_y}}{{v_x}})$theta, equals, tangent, start superscript, minus, 1, end superscript, left parenthesis, start fraction, v, start subscript, y, end subscript, divided by, v, start subscript, x, end subscript, end fraction, right parenthesis, the fact that we put $v_y$v, start subscript, y, end subscript on top as the opposite side and the $v_x$v, start subscript, x, end subscript 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 $v_x$v, start subscript, x, end subscript is positive, the vector points rightward. If the horizontal component, $v_x$v, start subscript, x, end subscript is negative, the vector points leftward.

Again, asumming we have selected right/up as the positive directions, if the vertical component $v_y$v, start subscript, y, end subscript is positive, the vector points upward. If the vertical component $v_y$v, start subscript, y, end subscript is negative, the vector points downward.

So, for example, if the components of a vector are $v_x=-12 \text{ m/s}$v, start subscript, x, end subscript, equals, minus, 12, start text, space, m, slash, s, end text and $v_y=10 \text{ m/s}$v, start subscript, y, end subscript, equals, 10, start text, space, m, slash, s, end text, the vector must point leftward—because $v_x$v, start subscript, x, end subscript is negative—and up—because $v_y$v, start subscript, y, end subscript is positive.

A soccer ball is kicked up and to the right at an angle of 30$^\circ$degrees with a speed of 24.3 m/s as seen below.

To find the vertical component of the velocity, we'll use $sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{v_y}{v}$s, i, n, theta, equals, start fraction, start text, o, p, p, o, s, i, t, e, end text, divided by, start text, h, y, p, o, t, e, n, u, s, e, end text, end fraction, equals, start fraction, v, start subscript, y, end subscript, divided by, v, end fraction. The hypotenuse is the magnitude of the velocity 24.3 m/s, $v$v, and the opposite side to the angle 30$^\circ$degrees is $v_y$v, start subscript, y, end subscript.

To find the horizontal component, we'll use $\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{v_x}{v}$cosine, theta, equals, start fraction, start text, a, d, j, a, c, e, n, t, end text, divided by, start text, h, y, p, o, t, e, n, u, s, e, end text, end fraction, equals, start fraction, v, start subscript, x, end subscript, divided by, v, end fraction.

An angry seagull is flying over Seattle with a horizontal component of velocity $v_x=14.6 \text{ m/s}$v, start subscript, x, end subscript, equals, 14, point, 6, start text, space, m, slash, s, end text and a vertical component of velocity $v_y=-8.62 \text{ m/s}$v, start subscript, y, end subscript, equals, minus, 8, point, 62, start text, space, m, slash, s, end text.

We'll use the Pythagorean theorem to find the magnitude of the total velocity vector.

To find the angle, we'll use the definition of $\text{tangent}$start text, t, a, n, g, e, n, t, end text, but since we now know $v$v, we could have used $\text{sine}$start text, s, i, n, e, end text or $\text{cosine}$start text, c, o, s, i, n, e, end text.

Since the vertical component is $v_y=-8.62 \text{ m/s}$v, start subscript, y, end subscript, equals, minus, 8, point, 62, start text, space, m, slash, s, end text, we know the vector is directed down, and since $v_x=14.6 \text{ m/s}$v, start subscript, x, end subscript, equals, 14, point, 6, start text, space, m, slash, s, end text, we know the vector is directed right. So, we will draw the vector in the fourth quadrant.

The seagull is moving $17.0 \text{ m/s}$17, point, 0, start text, space, m, slash, s, end text at an angle of $30.6^\circ$30, point, 6, degrees below the horizontal.