3.15 Kinematics (HL)

1. Equations of Motion

Vector equations are practically applied in physics to geometrically track the displacement motion of objects traveling at constant velocities along linear trajectories.

A body moving uniformly constructs a position vector function dependent entirely upon time ($t$):

$\vec{r} = \vec{a} + t\vec{v}$

$\vec{a}$ denotes the initial physical starting position evaluated strictly at time $t=0$.
$\vec{v}$ dictates the velocity vector. This represents exact directional displacement covered per single standardized unit of time.
$|\vec{v}|$ mathematically represents the physical speed (a pure scalar magnitude).

Critical Distinction: In geometric line equations, a direction vector $\vec{b}$ can be arbitrarily multiplied by any scalar without altering the continuous line. In kinematics, the velocity vector $\vec{v}$ cannot be scaled, because altering it functionally manipulates the object's explicit physical speed.

EXAMPLE 1 (Evaluating Constant Velocity)

A body operates under the motion profile $\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + t\begin{pmatrix} 4 \\ 3 \end{pmatrix}$ tracking meters per second.

Initial position coordinate traces to $(1,2)$, mapping a distance of $\sqrt{1^2 + 2^2} \approx 2.24\text{ m}$ from the origin boundary.
Following $1$ second ($t=1$), the structural position updates to $(5,5)$, situated $\sqrt{5^2 + 5^2} \approx 7.07\text{ m}$ away from the origin.
The velocity vector computes solidly as $\vec{v} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$.
The scalar speed equates to the absolute magnitude:

$|\vec{v}| = \sqrt{4^2 + 3^2} = \mathbf{5\text{ m/s}}$

EXAMPLE 2 (Constructing Velocity from Speed)

An object is traversing a 3D grid initiating from point $A(1,1,1)$, propelled in the absolute direction of vector $\vec{b} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ maintaining a strict constant speed of $15\text{ m/s}$. Formulate the dynamic motion equation.

Step 1: Determine the directional vector magnitude:
$|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3$

Step 2: Extract the geometric unit vector mapping a normalized length of $1$:
$\hat{b} = \dfrac{1}{3}\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$

Step 3: Multiply the normalized unit trace strictly by the required speed constraint ($15$) to formulate the true velocity vector:

$\vec{v} = 15\left[\dfrac{1}{3}\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}\right] = 5\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 10 \\ 10 \end{pmatrix}$

Step 4: The final kinematic relation calculates seamlessly as:

$\vec{r} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + t\begin{pmatrix} 5 \\ 10 \\ 10 \end{pmatrix}$

2. Geometric Paths versus Physical Collisions

Analyzing equations for two separate moving objects mandates differentiating cleanly between their geographical paths crossing and a synchronized physical impact.

Path Intersection (Geometrical): Do the lines conceptually cross? Evaluate the positional setup $\vec{r}_1 = \vec{r}_2$ utilizing distinctly separated time parameters (e.g., $t_1$ and $t_2$). If a valid pair of scalars emerges solving the system, the paths cross in space.
Physical Collision (Kinematical): Do the specific objects strike one another? Determine if the spatial paths intersect exactly when $\mathbf{t_1 = t_2}$. This mandates they arrive at the shared spatial coordinate at the identical absolute moment.

EXAMPLE 3 (Intersection vs Collision)

Two independent objects execute trajectories strictly modeled by:

$\vec{r}_1 = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} + t_1\begin{pmatrix} 5 \\ 4 \\ 3 \end{pmatrix} \quad \text{and} \quad \vec{r}_2 = \begin{pmatrix} 4 \\ 4 \\ 1 \end{pmatrix} + t_2\begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}$

Do their paths geometrically intersect, and do the bodies undergo a physical collision?

Step 1: Check Path Intersection
Set up the parametric system utilizing isolated variables ($t_1$ and $t_2$):

$$\begin{aligned} 3 + 5t_1 &= 4 + 3t_2 \implies 5t_1 - 3t_2 = 1 \\ 2 + 4t_1 &= 4 + 2t_2 \implies 4t_1 - 2t_2 = 2 \\ 1 + 3t_1 &= 1 + 2t_2 \implies 3t_1 - 2t_2 = 0 \end{aligned}$$

Resolving the bottom two equations produces precisely $\mathbf{t_1 = 2}$ and $\mathbf{t_2 = 3}$.
Inserting these values strictly back into the top equation mathematically validates the condition ($5(2) - 3(3) = 1 \implies 1 = 1$). Because the equations resolve perfectly, the paths do intersect geometrically at the coordinate $(13, 10, 7)$.

Step 2: Check Physical Collision
Although the lines logically cross, we observe that the arrival times are disconnected:

Object 1 reaches $(13, 10, 7)$ at $\mathbf{t_1 = 2\text{ s}}$.
Object 2 reaches $(13, 10, 7)$ at $\mathbf{t_2 = 3\text{ s}}$.

Since $\mathbf{t_1 \neq t_2}$, they strike the shared spatial node at entirely separated temporal moments. This confirms a geometric path intersection but conclusively rejects a physical collision.