Exhaustive Study Notes: 2D and 3D Vector Kinematics

Administrative Updates and Corrections

• Wednesday Lecture Problem Correction:

  • The solution presented on the slides regarding the time intervals was correct, but contained a typographical error in the integration/interval bounds.

  • The bound was incorrectly listed on the slide as $\Delta x$ from $1$ to $2$, but it should have been listed as $\Delta x$ from $0$ to $1$.

  • The final ratio statement remains correct: the time interval from $t = 1\,\text{s}$ to $t = 2\,\text{s}$ (denoted $\Delta t_{1 \to 2}$) is three times ($3\times$) the time interval from $t = 0\,\text{s}$ to $t = 1\,\text{s}$ (denoted $\Delta t_{0 \to 1}$).

  • During the real-time substitution working of the problem, the substitute professor mistakenly evaluated position $x$ at the end of $2\,\text{s}$ (yielding $x = 4\,\text{m}$) and compared it to $x = 1\,\text{m}$ at $t = 1\,\text{s}$, concluding a factor of four ($4\times$). Subtracting the position at $t = 1\,\text{s}$ ($x = 1\,\text{m}$) from position at $t = 2\,\text{s}$ ($x = 4\,\text{m}$) gives the correct displacement $\Delta x_{1 \to 2} = 3\,\text{m}$.

• Distance versus Displacement Clarification:

  • Distance is a scalar quantity representing path length, whereas displacement is a vector representing net change in position.

  • In a setup where a cart moves over a path, specifying displacement of $30\,\text{m}$ creates ambiguity because both the cart at an earlier time and at a higher point have an identical displacement of $30\,\text{m}$ relative to the origin. Under displacement alone, velocity cannot be uniquely solved as $+6\,\text{m/s}$ versus $-6\,\text{m/s}$.

  • Specifying a distance of $30\,\text{m}$ uniquely singles out the upward path spanning from $t = 0\,\text{s}$ to $t = 3.33\,\text{s}$ (or $3\frac{1}{3}\,\text{s}$), yielding a positive velocity of $+6\,\text{m/s}$.

  • Selecting the alternative velocity solution of $-6\,\text{m/s}$ corresponds to a total trajectory distance of $50\,\text{m}$.

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Position and Displacement Vectors in Multi-Dimensional Space

• Spatial Dimension Axes Convention:

  • Motion is extended from one dimension ($1\text{D}$) to two ($2\text{D}$) and three ($3\text{D}$) spatial dimensions.

  • Standard Cartesian spatial orientation:

    • $x$

      • Points horizontally to the right.

    • $y$

      • Points directly into the page/screen.

    • $z$

      • Points vertically upward.

• Position Vector Definition:

  • The spatial position of a point $P$ in 3D space at time $t$ is defined by the position vector $\mathbf{r}(t)$:

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

*   Basis unit vectors represent direction vectors of unit magnitude ($1$):
    *   $\hat{i}$
        *   Unit vector along the $x$-axis (equivalent notation: $\hat{x}$).
    *   $\hat{j}$
        *   Unit vector along the $y$-axis (equivalent notation: $\hat{y}$).
    *   $\hat{k}$
        *   Unit vector along the $z$-axis (equivalent notation: $\hat{z}$).

• Displacement Vector Definition:

  • The displacement vector $\Delta \mathbf{r}$ between an initial position at time $t_1$ and a final position at time $t_2$ is the vector difference:

$\Delta \mathbf{r} = \mathbf{r}(t_2) - \mathbf{r}(t_1)$

*   Vector subtraction is executed graphically by flipping the direction of $\mathbf{r}(t_1)$ and joining it head-to-tail with $\mathbf{r}(t_2)$.
*   Displacement measures net vector change between endpoints; it is independent of the curved trajectory length traveled (distance).

• Worked Example: Fly Trajectory Calculation:

  • An insect moves from initial position vector $\mathbf{r}_1$ to final position vector $\mathbf{r}_2$:

    • $\mathbf{r}_1 = 2\hat{i} - 1\hat{j} + 0\hat{k}\,\text{m}$

    • $\mathbf{r}_2 = -3\hat{i} + 4\hat{j} + 2\hat{k}\,\text{m}$

  • Displacement Vector Component Evaluation:

$\Delta x = -3 - 2 = -5\,\text{m}$ $\Delta y = 4 - (-1) = 5\,\text{m}$ $\Delta z = 2 - 0 = 2\,\text{m}$ $\Delta \mathbf{r} = -5\hat{i} + 5\hat{j} + 2\hat{k}\,\text{m}$

*   **Magnitude of Displacement Vector:**

$|\Delta \mathbf{r}| = \sqrt{(-5)^2 + 5^2 + 2^2} = \sqrt{25 + 25 + 4} = \sqrt{54}\,\text{m}$

*   The magnitude $|\Delta \mathbf{r}| = \sqrt{54}\,\text{m}$ represents the direct straight-line distance between initial and final points, which is strictly less than or equal to the actual curved path length.

Derivatives of Vectors and Velocity

• Instantaneous Velocity Vector Definition:

  • The instantaneous velocity vector $\mathbf{v}(t)$ is the time derivative of the position vector $\mathbf{r}(t)$:

$\mathbf{v}(t) = \lim_{\Delta t \to 0} \frac{\mathbf{r}(t + \Delta t) - \mathbf{r}(t)}{\Delta t} = \frac{d\mathbf{r}}{dt}$

*   The velocity vector points tangent to the spatial path at every instant in time, indicating the direction of continuous motion.

• Speed Definition:

  • Speed $v$ is a scalar quantity equal to the magnitude of the instantaneous velocity vector:

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

*   Taking the vector magnitude discards directional information and isolates rate of motion.

• Component-Wise Vector Differentiation:

  • Differentiating $\mathbf{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k}$ with respect to time involves taking derivatives of component scalars:

$\mathbf{v}(t) = \frac{dx(t)}{dt}\hat{i} + \frac{dy(t)}{dt}\hat{j} + \frac{dz(t)}{dt}\hat{k}$

*   The time derivatives of unit basis vectors are identically zero ($\frac{d\hat{i}}{dt} = \frac{d\hat{j}}{dt} = \frac{d\hat{k}}{dt} = 0$) because fixed coordinate basis vectors do not change magnitude or orientation over time.
*   Individual velocity scalar components are defined as:

$v_x(t) = \frac{dx(t)}{dt}$ $v_y(t) = \frac{dy(t)}{dt}$ $v_z(t) = \frac{dz(t)}{dt}$ $\mathbf{v}(t) = v_x(t)\hat{i} + v_y(t)\hat{j} + v_z(t)\hat{k}$

• Average Velocity Vector:

  • Average velocity $\mathbf{v}_{\text{avg}}$ over finite time interval $\Delta t = t_2 - t_1$ is strictly determined by spatial endpoints:

$\mathbf{v}_{\text{avg}} = \frac{\mathbf{r}(t_2) - \mathbf{r}(t_1)}{t_2 - t_1} = \frac{\Delta \mathbf{r}}{\Delta t}$

*   Average velocity ignores path history. Moving along a complex path and returning to the exact starting point yields $\mathbf{v}_{\text{avg}} = 0$, whereas average speed over that path is non-zero.

Acceleration Vector and Motion Independence

• Independence of Perpendicular Spatial Motion:

  • Motion along perpendicular axes ($x$, $y$, and $z$) operates completely independently.

  • A projectile launched horizontally off a height experiences horizontal motion with constant horizontal velocity ($v_x = \text{constant}$) while simultaneously accelerating vertically downward under gravity ($a_y = -g$).

  • The independent component equations share only one scalar variable: time $t$.

• Acceleration Vector Definition:

  • The acceleration vector $\mathbf{a}(t)$ is the time derivative of the velocity vector or second derivative of position:

$\mathbf{a}(t) = \lim_{\Delta t \to 0} \frac{\mathbf{v}(t + \Delta t) - \mathbf{v}(t)}{\Delta t} = \frac{d\mathbf{v}}{dt} = \frac{d^2\mathbf{r}}{dt^2}$

*   Component formulation:

$a_x(t) = \frac{dv_x}{dt} = \frac{d^2x}{dt^2}$ $a_y(t) = \frac{dv_y}{dt} = \frac{d^2y}{dt^2}$ $a_z(t) = \frac{dv_z}{dt} = \frac{d^2z}{dt^2}$ $\mathbf{a}(t) = a_x(t)\hat{i} + a_y(t)\hat{j} + a_z(t)\hat{k}$

• Causes of Non-Zero Acceleration:

  • Acceleration occurs if the magnitude of velocity (speed) changes, OR if the spatial direction of the velocity vector changes.

  • Example: Planets orbiting in circular trajectories at constant scalar speed possess continuous inward acceleration because the orientation of the velocity vector continually rotates.

Kinematic Equations for Constant Acceleration in 2D

• Independent Directional Kinematic Equations:

  • $x$--Component Equations:

$v_{\text{avg}, x} = \frac{\Delta x}{\Delta t}$ $v_x(t) = v_{0x} + a_x t$ $x(t) = x_0 + v_{0x} t + \frac{1}{2}a_x t^2$ $v_x^2 = v_{0x}^2 + 2a_x(x - x_0)$

*   **$y$--Component Equations:**

$v_{\text{avg}, y} = \frac{\Delta y}{\Delta t}$ $v_y(t) = v_{0y} + a_y t$ $y(t) = y_0 + v_{0y} t + \frac{1}{2}a_y t^2$ $v_y^2 = v_{0y}^2 + 2a_y(y - y_0)$

• Worked Example: Drone Trajectory Evaluation:

  • A drone at $t = 0\,\text{s}$ has initial position $\mathbf{r}_0$, initial velocity $\mathbf{v}_0 = 3\hat{i} + 4\hat{j}\,\text{m/s}$, and constant acceleration $\mathbf{a} = 1\hat{i} - 2\hat{j}\,\text{m/s}^2$

  • Method 1: Component Calculation at $t = 3\,\text{s}$:

    • $x$-velocity component:

$v_x(3) = v_{0x} + a_x t = 3 + (1)(3) = 6\,\text{m/s}$

    *   $y$-velocity component:

$v_y(3) = v_{0y} + a_y t = 4 + (-2)(3) = 4 - 6 = -2\,\text{m/s}$

    *   Combined velocity vector:

$\mathbf{v}(3) = 6\hat{i} - 2\hat{j}\,\text{m/s}$

    *   Position components at $t = 3\,\text{s}$:

$x(3) = x_0 + (3)(3) + \frac{1}{2}(1)(3)^2 = x_0 + 9 + 4.5 = x_0 + 13.5\,\text{m}$ $y(3) = y_0 + (4)(3) + \frac{1}{2}(-2)(3)^2 = y_0 + 12 - 9 = y_0 + 3\,\text{m}$ $\mathbf{r}(3) = (x_0 + 13.5)\hat{i} + (y_0 + 3)\hat{j}\,\text{m}$

*   **Method 2: Direct Vector Notation Calculation:**

$\mathbf{v}(t) = \mathbf{v}_0 + \mathbf{a}t$ $\mathbf{v}(3) = (3\hat{i} + 4\hat{j}) + (1\hat{i} - 2\hat{j})(3) = (3 + 3)\hat{i} + (4 - 6)\hat{j} = 6\hat{i} - 2\hat{j}\,\text{m/s}$ $\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2}\mathbf{a}t^2$ $\mathbf{r}(3) = \mathbf{r}_0 + (3\hat{i} + 4\hat{j})(3) + \frac{1}{2}(1\hat{i} - 2\hat{j})(3)^2 = \mathbf{r}_0 + 13.5\hat{i} + 3\hat{j}\,\text{m}$

Decomposing Vectors on Inclined Planes

• Incline Plane Acceleration Problem:

  • An object accelerates down an incline angled at $15^\circ$ below the horizontal with acceleration magnitude $a = 2.1\,\text{m/s}^2$

  • Standard coordinate alignment: positive $x$ points horizontally right, positive $y$ points vertically up.

  • Angle measured from positive $x$-axis is $\theta = -15^\circ$ (clockwise rotation below horizontal).

  • Acceleration Vector Decomposition:

$a_x = a \cos(-15^\circ) = 2.1 \cos(15^\circ)\,\text{m/s}^2$ $a_y = a \sin(-15^\circ) = -2.1 \sin(15^\circ)\,\text{m/s}^2$ $\mathbf{a} = 2.1 \cos(15^\circ)\hat{i} - 2.1 \sin(15^\circ)\hat{j}\,\text{m/s}^2$

*   **Sanity Checking Vector Signs:**
    *   Because spatial movement is directed downward and rightward, the vertical component $a_y$ must evaluate to a negative number when upward is designated as positive $y$.

Questions & Discussion

• Question on Basis Vector Terminology:

  • Question: Can basis vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$ be substituted with $\hat{x}$, $\hat{y}$, and $\hat{z}$?

  • Response: Yes, $\hat{x}$, $\hat{y}$, and $\hat{z}$ represent identical physical unit directions. However, $\hat{i}$, $\hat{j}$, $\hat{k}$ is standard notation across standard physics literature.

• Question on Formula Memorization for Exams:

  • Question: Are students required to memorize fundamental kinematic equations for exams, or will equation sheets be provided?

  • Response: Fundamental kinematic equations must be memorized thoroughly. Equation sheets provided during exams will only include complex or esoteric formulas.

• Discussion on Rotating Coordinate Axes (Incline Motion):

  • Student Proposal (Keenan): Rotating axes such that the primary $x$-axis aligns directly down the incline slope simplifies the acceleration vector to $\mathbf{a} = 2.1\hat{i}'\,\text{m/s}^2$

  • Analysis: Axis rotation simplifies acceleration, but requires transforming initial position $\mathbf{r}_0$ and velocity $\mathbf{v}_0$ vectors into the rotated frame. Decomposing the acceleration vector into standard unrotated Cartesian coordinates is typically less prone to coordinate conversion errors.

• Clarification on Angle Signs (Damien):

  • Discussion: Pointed out that because acceleration acts downward relative to horizontal positive $y$, the angle input must be $-15^\circ$ (or specified as $-2.1 \sin(15^\circ)\hat{j}$), confirming that vertical acceleration $a_y$ is negative.