Jump to a key chapter

Instead of seeing the cars move only along one straight line, you'll see them move forward, backward, and turning in various directions. Such motion can no longer be described using a single line, as it's two-dimensional, so a y-axis is introduced to effectively describe it. The third dimension can be visualized if we imagine one of the roads going up a hill, some cars would be moving up the hill and some down. This up and down motion would be along the third z-axis. In this article, we'll learn more about motion in two or three dimensions.

## Definition of Motion in Two Dimensions

Although we observe things moving daily, rarely do we dwell on the question "how many dimensions is this motion occurring in?" One of the simplest examples of movement is an object moving linearly in one dimension. Let's go back to the highway example. A car moving along the x-axis represents motion in one dimension. If that same car was moving in one direction with a constant velocity while accelerating in another direction, it would be **two-dimensional motion.**

**Two-dimensional motion** is motion that takes place along two different directions at the same time and requires at least two co-ordinates to be described mathematically.

Many of the moving objects we observe and experience every day do not travel in straight lines, and their motion is best described as two-dimensional. In one dimension, we analyze the motion of objects in a straight line by studying *displacement*, *velocity*, and *acceleration*. Now, how can we extend what we have learned about motion in one dimension to two-dimensional situations?

## Forces and Motions in Two Dimensions

To make a distinction between** **force and motion**,** let's define exactly what we mean by **force**.

**Force** is a vector quantity that describes the interactions between objects or systems.

Forces exerted on objects are always due to the interaction of that object with another object, like a push or a pull. An example of a force is throwing a baseball, as the process can be simplified into an object being pushed through the air. Similarly, when friction acts between two surfaces, we must consider both, the frictional force that is parallel to the surface and the normal force perpendicular to it. These are best described using x and y co-ordinates in two dimensions.

Two great examples of two-dimensional motion are projectile and circular, where the relevant forces are gravitational and centripetal, respectively.

### Projectile Motion

Projectile motion is a special case of two-dimensional motion.

**Projectile motion** is the motion of an object that has zero acceleration in one dimension and non-zero acceleration in the second dimension.

When analyzing projectile motion, it is critical to remember that it is always best to break the motion up into two components: one along the horizontal axis, and one along the vertical axis.

In projectile motion, vertical and horizontal components are independent of each other.

When a projectile is launched at an angle, the object is moving in both the x and y directions simultaneously. These can be expressed using vectors, where the addition will result in a net vector. Most of the time, projectile motion is modeled without including the effects of air resistance; hence it can be ignored.

When neglecting air resistance, the horizontal velocity component of a projectile has no net force acting upon it, while the vertical component is affected only by the force of gravity, meaning the acceleration term \(a_y\) can be replaced with the acceleration due to gravity \(g\).

### Uniform Circular Motion

Another common example of two-dimensional motion is **uniform circular motion**.

**Uniform circular motion** is the movement of an object along a circular path of a constant radius with constant speed.

Even though the speed is constant, the velocity is not. This is because the direction of the velocity vector changes from point to point along the path it travels. The fact that the velocity changes means that the acceleration is non-zero.

In circular motion, there is always a *centripetal acceleratio**n* whose magnitude can be expressed as \(a_\mathrm{c}\),

\[a_\mathrm{c}=\frac{v^2}{r},\]

where \(r\) is the radius of the circle. As a result, there is a *centripetal force* that points to the center of the circle defined by the object's motion. This force changes the direction of the velocity vector of the object, but not its magnitude. Finally, the time it takes to complete a full revolution is known as a period \(T\) and is given by the equation

\[T=\frac{2 \pi r}{v}.\]

A few good examples of circular motion are a Ferris wheel, or a satellite orbiting a planet.

## Formulas for Motion in Two Dimensions

Before we list all the relevant motion equations for two dimensions, let's recall the main equations of motion in one dimension.

Table 1 - Kinematic equations for motion in one dimension with constant acceleration.

Formula | Missing variable | Variables |

\(v=v_0+at\) | \(\Delta x\) | \(v\) - velocity\(v_0\) - initial velocity\(a\) - acceleration\(t\) - time |

\(x -x_0=v_0t+\frac{1}{2}at^2\) | \(v\) | \(x\) - position\(x_0\) - initial position\(v_0\) - initial velocity\(a\) - acceleration\(t\) - time |

\(v^2=v^2_0+2a(x-x_0)\) | \(t\) | \(v\) - velocity\(v_0\) - initial velocity\(a\) - acceleration\(x\) - position\(x_0\) - initial position |

Motion in two dimensions is described by the same equations as the ones compiled in Table 1 above, the only difference is in notation. In two-dimensional motion we consider two directions (x and y) so the notation for position becomes:

\[ \Delta \vec{r}=\vec{r}_2-\vec{r}_1,\]

where \( \vec{r}_1\) and \( \vec{r}_2\) are vectors from the origin to the points \( A\) and \( B\) in Figure 2. Each of these vectors has x and y components.

In this case, the average velocity is the displacement vector \( \overrightarrow{A B}\) divided by the time it takes for the object to move from \( A\) to \( B\). Instantaneous velocity, on the other hand, can be expressed as

\[\vec{v}= \frac{\mathrm{d}\vec{r}}{\mathrm{d}t},\]

where the x component is equal to

\[v_x= \frac{\mathrm{d}x}{\mathrm{d}t}.\]

The instantaneous acceleration is the change in velocity over time:

\[\vec{a}= \frac{\mathrm{d}\vec{v}}{\mathrm{d}t}.\]

Since it is not convenient to do calculations with vectors, to solve problems in two-dimensional motion, we use the components of the vectors. Then the vector equations turn into algebraic equations, the same as the equations used in one-dimensional motion, which are easier to solve using elementary mathematical operations. Following that logic, all the equations in Table 1 can be split into x and y components:

\begin{align} v_x&=v_{0 x}+a_x t \\ v_y&=v_{0 y}+a_y t \end{align}

when calculating velocity,

\begin{align} x-x_0&=v_{0 x} t+\frac{1}{2} a_x t^2 \\ y-y_0&=v_{0 y} t+\frac{1}{2} a_y t^2, \end{align}

for obtaining the position components, and

\begin{align} v_x^2&=v_{0 x}^2 +2 a_x(x-x_0) \\ v_y^2&=v_{0 y}^2 +2 a_y(y-y_0). \end{align}

Breaking up expressions into components turns a two-dimensional motion problem into a system of two equations with two unknowns.

Note that the time \(t\) in the equations above is the same for the two components.

It's easier to understand these equations through application in example problems.

## Examples of Motion in Two Dimensions

Let's look at an example problem of projectile motion in two dimensions.

A ball is launched at an angle of \(33^\circ\) above the horizontal, as pictured in Figure 3. If it has an initial speed of \(20 \, \frac{\mathrm{m}}{\mathrm{s}}\), calculate:

- how much time does the ball spend in the air,
- the maximal height of the projection, and
- the maximum horizontal displacement.

Use \(9.8 \, \frac{\mathrm{m}}{\mathrm{s^2}}\) for the acceleration due to gravity.

**Answer**:

First, we must find the time of flight \(t\). This quantity depends only on the vertical component of the projectile. Before the launching and after the ball reaches the ground, it will have a \(\Delta y = 0\), so we can set it to zero in the following equation:

\begin{align} \Delta y &= v_{0y}t + \frac{1}{2}a_yt^2 \\ 0&=v_{0y}t + \frac{1}{2}a_yt^2. \end{align}

Now we can use the fact that acceleration in the vertical direction is equal to the acceleration due to gravity \(g\) and the trigonometric properties of the triangle formed by \(v_{0y}\) and \(v_{0}\) to obtain the following expression:

\begin{align} (v_0\sin \theta)t+\frac{1}{2}(-g)t^2 &= 0 \\ \frac{1}{2}gt^2 - (v_0 \sin \theta)t &=0 \\ t \left ( \frac{1}{2}gt - v_0 \sin \theta \right )&=0. \end{align}

Before the projectile is launched \(t=0\), however, once it lands it will be non-zero, so we can simplify the expression to

\[\frac{1}{2}gt - v_0 \sin \theta=0 .\]

Finally, we just plug in the known values:

\begin{align}t&=\frac{2v_0\sin \theta}{g} \\ t&=\frac{2\left( 20 \, \frac{\mathrm{m}}{\mathrm{s}} \right )(\sin 33^\circ)}{9.8 \, \frac{\mathrm{m}}{\mathrm{s^2}}} \\ t&=2.2 \, \mathrm{s}. \end{align}

Second, we are asked to calculate the maximum height of the projection \(\Delta y\), which can be done by utilizing

\[v^2=v^2_0+2a(x-x_0).\]

In this case, we care for the y component, so we can alter the expression to

\[v^2_y=v^2_{0y}+2a(y-y_0).\]

As the ball reaches the highest point \(v_y\), its velocity will be zero, resulting in the following simplification:

\begin{align} 0 &= v^2_{0y}+2a(\Delta y) \\ -v^2_{0y}&=2a(\Delta y) \\ \Delta y&= \frac{-v^2_{0y}}{2a}. \end{align}

Once again, we can use trigonometric properties to express the top term into our known variables and calculate the maximum height of the ball:

\begin{align} y_\mathrm{max}&= \frac{-v^2_{0y}}{2a} \\ &= \frac{-(v_0 \sin \theta)^2}{2(-g)} \\ &= \frac{\left( 20 \, \frac{\mathrm{m}}{\mathrm{s}} \right )^2 (\sin 33^\circ)^2}{2\cdot( 9.8 \, \frac{\mathrm{m}}{\mathrm{s^2}})} \\ &= 6.1 \, \mathrm{m}. \end{align}

Finally, let's calculate the maximum horizontal displacement \(\Delta x \). We know that velocity is

\[v_x=\frac{\Delta x}{t},\]

which can be rearranged into

\[\Delta x = v_x t.\]

We can use the previously calculated value for time, and the x component of the velocity term can be expressed using trigonometric properties and plugged into the displacement equation:

\begin{align} \Delta x &= (v_0\cos \theta) t \\ &= \, \left ( 20 \, \frac{\mathrm{m}}{\mathrm{s}}\right ) (\cos 33^\circ)(2.2 \, \mathrm{s})\\ &=37\, \mathrm{m}. \end{align}

## Motion in Three Dimensions

Just like converting from one to two-dimensional motion, objects in three dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components. To achieve that, a third axis is introduced, therefore now everything is described using x, y, and z coordinates.

Unit vector notation is applied to represent vectors as the sum of their constituent components in each direction. The usual coordinates of x, y, and z, are represented by \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\), respectively. For instance, a vector \(\vec{r}\) in a three dimension space can be written as

\begin{align} \vec{r}&=(A, B, C) \\ \vec{r}&=A\hat{i}+B\hat{j}+C\hat{k}. \end{align}

Unit vector notation can be applied in all dimensions, however, most of the time there is no need to complicate expressions in one or two dimensions.

An example of three-dimensional motion is the motion of a gyroscope. More on how to qualitatively describe the motion of a particle in three-dimensional space will be covered in later Mechanics topics such as Electricity and Magnetism.

## Motion in Two Dimensions - Key takeaways

- Two-dimensional motion is motion that takes place in two different directions (or coordinates) at the same time.
- Two examples of two-dimensional motion are projectile and circular, where the relevant forces are gravitational and centripetal, respectively.
- Projectile motion is the motion of an object that has zero acceleration in one dimension and non-zero acceleration in the second dimension.
- Uniform circular motion is the movement of an object along a circular path of a constant radius with constant speed.
- Motion in two or three dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components.
- Unit vector notation is applied to represent vectors as the sum of their constituent components in each direction.
- Velocity and acceleration may be different in each dimension and may be non-uniform.
- Motion in one dimension may be changed without causing a change in a perpendicular dimension.

## References

- Fig. 1 - (https://unsplash.com/photos/xTAF3D6K-SU) by CHUTTERSNAP (https://unsplash.com/@chuttersnap) on Unsplash is licensed by Public Domain.
- Fig. 2 - Two-dimensional motion vectors, StudySmarter Originals.
- Fig. 3 - Projectile motion example, StudySmarter Originals.
- Fig. 4 - Silver coloured gyroscope by Hugöl Hälpingston (https://unsplash.com/@hugoheppo) on Unsplash is licensed by Public Domain.

###### Learn with 13 Motion in Two Dimensions flashcards in the free StudySmarter app

We have **14,000 flashcards** about Dynamic Landscapes.

Already have an account? Log in

##### Frequently Asked Questions about Motion in Two Dimensions

What is motion in two dimensions?

Motion in two dimensions is motion that takes place in two different directions (or coordinates) at the same time.

Can there be motion in two dimensions?

Yes, there can be motion in two dimensions, if motion occurs in two different directions simultaneously.

What do you call a motion that moves in two dimensions?

A motion that moves in two dimensions can be projectile or circular motion.

What is the difference between one-dimensional and two-dimensional motion?

The difference between one-dimensional and two-dimensional motion is that in one dimension movement is happening only in one direction, while in two dimensions objects are moving in two different directions.

How would you describe a position in two dimensions?

The position in two-dimensions can be described by introducing a coordinate system, where x represents the horizontal position and y gives the vertical position.

##### About StudySmarter

StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

Learn more