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!
