Vector Applications to Kinematics in Math AI
Modeling Linear Motion with Constant Velocity
In the realm of vector applications to kinematics, we begin by examining linear motion with constant velocity in two and three dimensions. This fundamental concept forms the basis for more complex motion analysis.
Two-Dimensional Motion
In a two-dimensional plane, we can represent the position of an object at any time $t$ using the vector equation:
$$\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}t$$
Where:
- $\mathbf{r}(t)$ is the position vector at time $t$
- $\mathbf{r}_0$ is the initial position vector
- $\mathbf{v}$ is the constant velocity vector
- $t$ is time
Consider a car moving with a constant velocity of 5 m/s east and 3 m/s north. If its initial position is at (2, 1), its position after 4 seconds can be calculated as:
$\mathbf{r}(4) = (2, 1) + (5, 3) \cdot 4 = (2, 1) + (20, 12) = (22, 13)$
The car will be at position (22, 13) after 4 seconds.
Three-Dimensional Motion
The same principle extends to three-dimensional space, where we add a z-component to our vectors:
$$\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}t = (x_0, y_0, z_0) + (v_x, v_y, v_z)t$$
NoteThe beauty of vector notation is that it allows us to treat multi-dimensional motion with the same simplicity as one-dimensional motion.
Finding Positions and Intersections
Using the vector equation of motion, we can solve various problems related to positions and intersections of moving objects.
Intersection of Paths
To find where two objects' paths intersect, we set their position vectors equal to each other and solve for time:
$$\mathbf{r}_1(t) = \mathbf{r}_2(t)$$
ExampleObject A starts at (0, 0) with velocity (2, 3) m/s, and object B starts at (10, 5) with velocity (-1, 1) m/s. To find their intersection:
$(0, 0) + (2, 3)t = (10, 5) + (-1, 1)t$
Solving this system of equations gives us the time and point of intersection.
Describing Paths
The path of an object can be described by eliminating the time parameter from the position vector equation. This often results in an equation relating x and y (and z in 3D).
ExampleFor a projectile launched with initial velocity $(v_0 \cos \theta, v_0 \sin \theta)$ and initial position $(0, 0)$, the path is described by:
$y = x \tan \theta - \frac{gx^2}{2v_0^2 \cos^2 \theta}$
This is the familiar parabolic path of projectile motion.
Closest Approach
Finding the time when two objects are closest to each other involves minimizing the distance between their position vectors. This is typically done by minimizing the square of the distance to avoid dealing with square roots.
$$d^2 = |\mathbf{r}_1(t) - \mathbf{r}_2(t)|^2$$
Differentiate with respect to t and set to zero to find the minimum.
Common MistakeStudents often forget that the closest approach may occur at a time when the objects have already passed each other. Always check if the solution makes physical sense!
Relative Position
The relative position of object B as seen from object A is given by the vector $\mathbf{AB}$:
$$\mathbf{AB} = \mathbf{r}_B - \mathbf{r}_A$$
This concept is crucial for understanding relative motion and solving problems involving multiple moving objects.
Modeling Motion with Variable Velocity
While constant velocity motion provides a good starting point, real-world scenarios often involve variable velocity. In two dimensions, this can be represented as:
$$\mathbf{v}(t) = \langle v_x(t), v_y(t) \rangle$$
Examples of Variable Velocity
A common example is motion with constant acceleration, where velocity changes linearly with time:
$$\mathbf{v}(t) = \mathbf{v}_0 + \mathbf{a}t$$
ExampleConsider an object with initial velocity $\langle 7, 6 \rangle$ m/s and acceleration $\langle 0, -4 \rangle$ m/s². Its velocity at time t is:
$\mathbf{v}(t) = \langle 7, 6-4t \rangle$
To find the position, we integrate this velocity:
$\mathbf{r}(t) = \int \mathbf{v}(t) dt = \langle 7t, 6t-2t^2 \rangle + \mathbf{C}$
Where $\mathbf{C}$ is the constant of integration, representing the initial position.
Special Cases: Projectile and Circular Motion
Projectile Motion
Projectile motion is a special case of two-dimensional motion under constant acceleration (gravity). The velocity components are:
$$v_x(t) = v_0 \cos \theta$$ $$v_y(t) = v_0 \sin \theta - gt$$
Where $\theta$ is the launch angle and g is the acceleration due to gravity.
Circular Motion
Circular motion involves constant speed but changing direction. In parametric form:
$$x(t) = R \cos(\omega t)$$ $$y(t) = R \sin(\omega t)$$
Where R is the radius and $\omega$ is the angular velocity.
TipWhen dealing with circular motion, it's often helpful to use polar coordinates instead of Cartesian coordinates.
Time Shifts and Phase Shifts
A time shift in a motion equation is represented by replacing t with (t-a):
$$f(t-a)$$
This indicates that the motion is shifted a units in time.
ExampleIf an object's position is given by $x(t) = \cos(t)$, then $x(t-\pi/2) = \cos(t-\pi/2)$ represents the same motion but starting a quarter cycle later.
This concept links directly to phase shifts in trigonometric functions (AHL 1.13) and is crucial in understanding periodic motion in kinematics (AHL 5.13).
NoteTime shifts are particularly important in analyzing systems with multiple moving parts or when comparing motions that start at different times.
Conclusion
Vector applications in kinematics provide a powerful tool for analyzing motion in two and three dimensions. By representing position, velocity, and acceleration as vectors, we can tackle complex motion problems with elegance and efficiency. The ability to model both constant and variable velocity scenarios, understand relative motion, and apply concepts like time shifts opens up a wide range of real-world applications, from predicting the paths of projectiles to analyzing the motion of celestial bodies.
TipWhen approaching kinematics problems, always start by clearly defining your coordinate system and writing out the vector equations of motion. This systematic approach will make even complex problems more manageable.
