3.15 KinematicsHL ONLY

1. Equations of Motion

Vector equations are used in physics to track the motion of objects moving at constant velocities along straight lines.

An object moving at a constant velocity has a position vector function dependent on time ($t$):

$\vec{r} = \vec{a} + t\vec{v}$
  • $\vec{a}$ is the initial position at time $t=0$.
  • $\vec{v}$ is the velocity vector, representing the displacement per unit of time.
  • $|\vec{v}|$ represents the speed (a scalar magnitude).

Note: In geometric line equations, a direction vector $\vec{b}$ can be multiplied by any scalar without changing the line. In kinematics, the velocity vector $\vec{v}$ cannot be scaled, because changing it changes the object's speed.

EXAMPLE 1 (Evaluating Constant Velocity)

An object moves according to the equation $\vec{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + t\begin{pmatrix} 4 \\ 3 \end{pmatrix}$, where distance is in meters and time is in seconds.

  • The initial position is $(1,2)$, which is a distance of $\sqrt{1^2 + 2^2} \approx 2.24\text{ m}$ from the origin.
  • At $t=1$, the position is $(5,5)$, which is $\sqrt{5^2 + 5^2} \approx 7.07\text{ m}$ from the origin.
  • The velocity vector is $\vec{v} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$.
  • The speed is the magnitude of the velocity vector:
    $|\vec{v}| = \sqrt{4^2 + 3^2} = \mathbf{5\text{ m/s}}$
x y O t=0 (1,2) t=1 (5,3) a v

EXAMPLE 2 (Constructing Velocity from Speed)

An object moves in 3D space starting from point $A(1,1,1)$ in the direction of vector $\vec{b} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ at a constant speed of $15\text{ m/s}$. Find the equation of motion.

Step 1: Find the magnitude of the direction vector:
$|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3$
Step 2: Find the unit vector:
$\hat{b} = \dfrac{1}{3}\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$
Step 3: Multiply the unit vector by the speed ($15$) to find the 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 equation of motion is:
$\vec{r} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + t\begin{pmatrix} 5 \\ 10 \\ 10 \end{pmatrix}$

2. Geometric Paths versus Physical Collisions

When analyzing equations for two moving objects, distinguish between their paths intersecting and a physical collision.

  • Path Intersection: The paths intersect if the lines cross. Set $\vec{r}_1 = \vec{r}_2$ using different parameters (e.g., $t_1$ and $t_2$). If the system has a solution, the paths intersect.
  • Physical Collision: A collision occurs if the objects are at the same place at the same time. This requires $t_1 = t_2$ at the intersection point.

EXAMPLE 3 (Intersection vs Collision)

Two objects move along the trajectories:

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

Determine if their paths intersect, and if the objects collide.

Step 1: Check Path Intersection
Set the equations equal using different parameters $t_1$ and $t_2$:
$$\begin{aligned} 3 + 5t_1 &= 4 + 3t_2 \implies 5t_1 - 3t_2 = 1 \\ 2 + 4t_1 &= 4 + 2t_2 \implies 4t_1 - 2t_2 = 2 \\ 1 + 3t_1 &= 1 + 2t_2 \implies 3t_1 - 2t_2 = 0 \end{aligned}$$
Solving the last two equations gives $t_1 = 2$ and $t_2 = 3$.
Substituting these into the first equation confirms the solution ($5(2) - 3(3) = 1$). The paths intersect at $(13, 10, 7)$.
Step 2: Check Physical Collision
The objects reach the intersection point at different times:
Object 1 reaches $(13, 10, 7)$ at $t_1 = 2\text{ s}$.
Object 2 reaches $(13, 10, 7)$ at $t_2 = 3\text{ s}$.
Since $t_1 \neq t_2$, the objects do not collide.