Lesson 2.5: Kinematics in Two Dimensions with Vectors

Introduction

In this lesson, we will explore the fascinating world of kinematics in two dimensions, examining how objects move in a plane. We will extend our understanding of one-dimensional motion with constant acceleration into two dimensions through the use of position, velocity, and acceleration vectors. By the end of this lesson, you will be able to:

Understand how to represent motion with vector notation.
Use calculus techniques to analyze vector functions of time.
Determine the speed and direction of motion using velocity vectors.
Apply constant-acceleration equations to two-dimensional motion.
Differentiate and integrate position and velocity vectors in $i$ and $j$ form.

Extension to Two-Dimensional Kinematics

Representing Motion Using Vectors

In one dimension, motion can be represented as a scalar quantity, but in two dimensions, we need to utilize vectors to define movement. A vector is defined by its magnitude (length) and direction.

Definition of Vectors

A two-dimensional vector can be expressed in terms of its components along the $x$ and $y$ axes. A position vector $\mathbf{r}(t)$ at time $t$ can be written as:

$$\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}$$

where:

$x(t)$ is the $x$-coordinate as a function of time.
$y(t)$ is the $y$-coordinate as a function of time.
$\mathbf{i}$ and $\mathbf{j}$ are unit vectors in the direction of the $x$ and $y$ axes, respectively.

Component Motion

When examining the motion of an object in two dimensions, it is useful to analyze the motion separately along each axis. This means we can use the equations of motion we learned for one-dimensional cases, but apply them to both axes independently.

Constant Acceleration in Two Dimensions

When an object is undergoing constant acceleration $\mathbf{a}$, along the $x$ and $y$ directions, we can express its position as:

For the x-direction:

$$x(t) = x_0 + v_{0x}t + \frac{1}{2}a_x t^2$$

For the y-direction:

$$y(t) = y_0 + v_{0y}t + \frac{1}{2}a_y t^2$$

where:

$x_0$ and $y_0$ are the initial positions.
$v_{0x}$ and $v_{0y}$ are the initial velocities.
$a_x$ and $a_y$ are the constant accelerations in the respective directions.

Worked Example 1

Example: A projectile is launched from the origin with an initial velocity of $\mathbf{v}_0 = 20\mathbf{i} + 30\mathbf{j}$ m/s. Suppose the object experiences a constant acceleration $\mathbf{a} = 0\mathbf{i} - 9.8\mathbf{j}$ m/s² (due to gravity). What is its position after 2 seconds?

Solution:

Break down the given quantities:

Initial position: $x_0 = 0$, $y_0 = 0$.
Initial velocity: $v_{0x} = 20$, $v_{0y} = 30$.
Acceleration: $a_x = 0$, $a_y = -9.8$.

Use the position equations:

For $x$:

$$x(2) = 0 + 20(2) + \frac{1}{2}(0)(2^2) = 40 \text{ m}$$

For $y$:

$$y(2) = 0 + 30(2) + \frac{1}{2}(-9.8)(2^2) = 60 - 19.6 = 40.4 \text{ m}$$

Therefore, the position vector after 2 seconds is:

$$\mathbf{r}(2) = 40\mathbf{i} + 40.4\mathbf{j} \text{ m}$$

Velocity and Acceleration Vectors

Velocity Vectors

The velocity vector $\mathbf{v}(t)$ is the derivative of the position vector with respect to time. Thus:

$$\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \frac{dx}{dt}\mathbf{i} + \frac{dy}{dt}\mathbf{j}$$

By applying the derivatives of the position functions:

$v_x(t) = v_{0x} + a_x t$
$v_y(t) = v_{0y} + a_y t$

Worked Example 2

Example: Continuing from the previous example, find the velocity of the projectile after 2 seconds.

Solution:

Using the velocity equations:

For $x$:

$$v_x(2) = 20 + 0 \cdot 2 = 20 \text{ m/s}$$

For $y$:

$$v_y(2) = 30 - 9.8 \cdot 2 = 30 - 19.6 = 10.4 \text{ m/s}$$

Therefore, the velocity vector at $t = 2$ seconds is:

$$\mathbf{v}(2) = 20\mathbf{i} + 10.4\mathbf{j} \text{ m/s}$$

Speed and Direction of Motion

The speed of an object is the magnitude of the velocity vector:

$$\text{Speed} = |\mathbf{v}(t)| = \sqrt{v_x^2 + v_y^2}$$

The direction can be found by calculating the angle $\theta$ with the horizontal axis:

$$\theta = \tan^{-1}\left(\frac{v_y}{v_x} ight)$$

Worked Example 3

Example: Find the speed and direction of the projectile after 2 seconds.

Solution:

Calculate the speed:

$$|\mathbf{v}(2)| = \sqrt{(20)^2 + (10.4)^2} = \sqrt{400 + 108.16} = \sqrt{508.16} \approx 22.5 \text{ m/s}$$

Calculate the direction:

$$\theta = \tan^{-1}\left(\frac{10.4}{20} ight) \approx 28.74°$$

Therefore, the speed is approximately $22.5$ m/s at an angle of $28.74°$ above the horizontal.

Calculus with Vector Functions

Differentiation of Position and Velocity Vectors

When dealing with vectors, differentiation must occur component-wise. If we have a position vector $\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}$, the derivative is:

$$\frac{d\mathbf{r}}{dt} = \frac{dx}{dt}\mathbf{i} + \frac{dy}{dt}\mathbf{j} = \mathbf{v}(t)$$

Integration of Velocity Vectors

Likewise, for integration, if you know the velocity vector, you obtain the position vector by integrating each component:

$$\mathbf{r}(t) = \int \mathbf{v}(t) dt = \int \left(v_{x}(t)\mathbf{i} + v_{y}(t)\mathbf{j} ight) dt$$

Worked Example 4

Example: Given a velocity vector defined as $\mathbf{v}(t) = (6t)\mathbf{i} + (8 - 4t)\mathbf{j}$ m/s, find the position vector $\mathbf{r}(t)$ if $\mathbf{r}(0) = 0$.

Solution:

Integrating component-wise:

For $x(t)$:

$$x(t) = \int 6t \, dt = 3t^2 + C_x$$

For $y(t)$:

$$y(t) = \int (8 - 4t) \, dt = 8t - 2t^2 + C_y$$

From the initial condition $\mathbf{r}(0) = 0$, we determine that $C_x = 0$ and $C_y = 0$. Thus:

Position vector becomes:

$$\mathbf{r}(t) = (3t^2)\mathbf{i} + (8t - 2t^2)\mathbf{j}$$

Conclusion

Kinematics in two dimensions provides a rich framework to analyze motion using vectors. We have learned to extend our knowledge from one-dimensional motion to analyze objects moving in a plane, breaking down their motion into components and evaluating it with calculus. By understanding how to differentiate and integrate vector functions, we can investigate motion in the $x$ and $y$ directions separately, find speed and direction, and ultimately describe the behavior of objects under various conditions.

Study Notes

Vectors are represented in terms of their components along the $x$ and $y$ axes.
The position vector in two dimensions is expressed as $\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}$.
The equations of motion can be applied independently to $x$ and $y$ directions for constant acceleration:
$x(t) = x_0 + v_{0x}t + \frac{1}{2}a_x t^2$
$y(t) = y_0 + v_{0y}t + \frac{1}{2}a_y t^2$
The velocity vector is found via differentiation: $\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}$.
Speed and direction can be derived from the velocity vector.
Calculus is used to find position and velocity vectors by integrating or differentiating components.