2D Motion and Projections

Systems thinking: what is a system?

• Systems biology referenced as fashionable about ~25 years ago; idea that a system can be anything, not just biology.
• A system consists of processes and the interfaces between those processes.
  • Metabolic pathways can be drawn as charts with arrows; these arrows represent interfaces where one process affects another (upregulation, downregulation, uptake, concentration changes).
  • You can describe how a system works by looking at interfaces, even if you don’t fully understand each individual process.
• Glitches in a system can be addressed by changing interfaces or reorganizing processes (systems engineering view).
  • This is the black box vs white box approach:
  • Black box: you don’t know internal mechanics; you focus on inputs/outputs and interfaces.
  • White box: you know how a process works; you can modify the process itself to improve performance.
• Practical takeaway: aim to improve the system, not just patch a single process.

Real-world examples of system glitches and fixes

• Blackboard to Canvas course transfer caused date misalignment; two subsystems involved: Canvas and Pearson (external grading/assignment platform).
• Observed behavior:
  • Canvas showed assignments with due dates wrong because data were fed from Pearson.
  • Canvas has its own assignment submission logic and late submission rules which can conflict with Pearson’s.
• Workarounds described (not recommended as ad hoc):
  • Do not manually fix each assignment date one by one (error-prone).
  • Instead, implement a feed that updates the dates at the source system or align rules across systems.
  • If Canvas shows questionable items, ignore them and rely on Pearson results, then override Canvas scores at the end.

Linking systems and avoiding clones

• In programming and data/workflow, avoid cloning/duplicating key components; reuse a single, well-maintained component (avoid drift in features).
• If you perfect an algorithm in one place, propagate the improvement everywhere instead of maintaining multiple copies.
• Analogy: smoothing algorithm for a chromatogram should be centralized so updates affect all software that uses it.

Hospital example: a system glitch and a better design

• Problem: patients spent extra time in admission due to delayed access to a Blood Pressure (BP) machine key.
  • A single nurse knew where the key was; delays occurred when this nurse was unavailable.
• Corrective action: replace workaround with a system that doesn’t depend on one person:
  • Use an electronic walker/BP access controlled by badges; doors open via badge scan.
  • Changes are easier to maintain when personnel changes are handled by updating a batch/employee identifier in a central database.
• Key lesson: redesign the system so the process is smooth and requires no single person’s intervention.

Management philosophy and ethical implications

• Do not blame individuals for system glitches; blame the process or the system itself.
• A well-designed system minimizes reliance on particular people and enables seamless collaboration.
• A good manager is often described as someone whose team appears to work magically without micro-management; the goal is a harmonious system where the team self-organizes.

Practical policy notes for course administration (Pearson vs Canvas)

• Instruction to students:
  • Use Pearson for homework/assignments; ignore Canvas entries if they look off.
  • Final scores will be overridden if necessary to reflect Pearson results.
• Policy on due dates and late submission (Pearson):
  • Each assignment has a due date.
  • Self-learning modules may close after the due date and may have fixed penalties (varies by Pearson type, not easily modifiable by the instructor).
  • For standard homework and pre/post lectures: penalties typically do not exceed 3% for late submissions.
• Practical classroom management tip: avoid manual edits across systems; focus on aligning the feeding of data between systems or overriding scores at the end if needed.

Two classic river-crossing scenarios: constant speed and the impact of current

• Setup conventions:
  • Two speeds: your own speed Vme (relative to water) and the river current speed Vr (along the river direction).
  • You cross a width L; x is downstream along the river; y is across the river.
  • When drawing vectors, velocity vectors are drawn aside from the body (not on the body) as a convention; trajectories are drawn as dotted lines parallel to the velocity direction.

Case 1: Constant speed across a river with current (two perpendicular motions)

• Given:
  • Your speed across (relative to water): $v_{me} = 0.4 ext{ m/s}$
  • River current: $v_r = 0.3 ext{ m/s}$
  • Width to cross: $L = 100 ext{ m}$
• Calculations:
  • Resultant velocity magnitude (Pythagoras since perpendicular components):
    $v<em>{tot} = \sqrt{v</em>{me}^2 + v_r^2} = \,\sqrt{0.4^2 + 0.3^2} = \,0.5 ext{ m/s}$
  • Time to cross (across component is what matters for time):
    $t = \frac{L}{v_{me}} = \frac{100}{0.4} = 250 ext{ s}$
  • Downstream drift during crossing (similar triangles; ratio of downstream to across equals ratio of river to cross speeds or, more geometrically, using similar triangles):
    $x = L \cdot \frac{v<em>r}{v</em>{me}} = 100 \cdot \frac{0.3}{0.4} = 75 ext{ m}$
• Summary:
  • Total speed magnitude: $v_{tot} = 0.5 ext{ m/s}$
  • Across-crossing time: $t = 250 ext{ s}$
  • Downstream drift: $x = 75 ext{ m}$

Case 2: Fight the current to land directly opposite the starting point (upstream angle)

• Setup:
  • You cannot swim faster than $v<em>{me} = 0.4 ext{ m/s}$, but you aim upstream so your across component is still vme in the y-direction while partially offset by the current.
• Calculations (simplified):
  • If you want to end up directly across (no downstream drift), the effective across-path velocity along the perpendicular to the current becomes the hypotenuse component with the current subtracted along the downstream direction, giving:
    $v<em>{tot} = \sqrt{v</em>{me}^2 - v_r^2} = \sqrt{0.16 - 0.09} \approx 0.265 ext{ m/s}$
  • Time to cross with this slower resultant:
    $t = \frac{L}{v_{tot}} = \frac{100}{0.265} \approx 377 ext{ s}$
• Takeaway:
  • Pushing directly across (no drift) while fighting the current increases the time spent in water; crossing downstream with the current is faster overall, but results in downstream landing.

Case 3: General landing with a specified downstream drift (geometry-based approach)

• Scenario:
  • Across distance A = 100 m; desired downstream landing offset D = 50 m.
• Geometry:
  • Angle to aim relative to across direction: $\alpha = \tan^{-1}\left( \frac{D}{A} \right) = \tan^{-1}(0.5) \approx 26.6^{\circ}$
  • Straight-line path length to landing: $H = \sqrt{A^2 + D^2} = \sqrt{100^2 + 50^2} \approx 112 \text{ m}$
  • Component relations:
  • The across-component of the total velocity relates to the angle: $v<em>{\perp} = v</em>{tot} \cos\alpha = v_{tot} \frac{A}{\sqrt{A^2 + D^2}}$
  • Note that $\cos\alpha = \frac{A}{\sqrt{A^2+D^2}} \approx 0.893$
• Key insight:
  • If you know the total speed, you can compute how far downstream you land; if you know the landing point and your total speed, you can infer the necessary angle. The two equations are consistent via similar triangles.
• Takeaway:
  • To land at a prescribed downstream offset, you typically adjust the aim angle so that the ratio of across to downstream components matches geometry; the exact total speed v_tot remains a design variable (your own power).

Ballistic trajectory (projectile motion) overview

• Setup:
  • Ignore air resistance for simplicity; define initial speed $v_0$ and launch angle $\alpha$; gravity $g \approx 9.8\ \text{m/s}^2$ downward.
  • Coordinates: x horizontal, y vertical; initial position (x0, y0).
• Kinematics:
  • Horizontal motion (no acceleration):
    $x(t) = x<em>0 + v</em>{0x} t, \ v<em>{0x} = v</em>0 \cos\alpha$
  • Vertical motion (constant downward acceleration):
    $y(t) = y<em>0 + v</em>{0y} t - \frac{1}{2} g t^2, \ v<em>{0y} = v</em>0 \sin\alpha$
• Key result: horizontal range (landing on same vertical level as launch, y = y0): $R = \frac{v_0^2}{g} \sin(2\alpha)$
  • Maximum range occurs at $\alpha = 45^{\circ}$, giving $R<em>{max} = \frac{v</em>0^2}{g}$
• Trajectory equation (path in the x-y plane):
  $y - y<em>0 = x \tan\alpha - \frac{g x^2}{2 v</em>0^2 \cos^2\alpha}$
• Alternative form by substituting $t = \tan\alpha$:
  • Start from $y - y<em>0 = \tan\alpha\,x - \frac{g x^2}{2 v</em>0^2 \cos^2\alpha}$ and use $\cos^2\alpha = \frac{1}{1 + \tan^2\alpha} = \frac{1}{1 + t^2}$ to obtain a quadratic in $t = \tan\alpha$:
    $a t^2 + b t + c = 0$
    with
    $a = \frac{g x^2}{2 v<em>0^2}, \quad b = -x, \quad c = (y - y</em>0) + \frac{g x^2}{2 v_0^2}$
  • Solve for the launch angle via
    $\tan\alpha = t = \frac{x \pm \sqrt{x^2 - 4ac}}{2a}$
• Ballistic notes:
  • There are two launch angles that can yield the same range (high vs low ballistic trajectories): because $\sin(2\alpha)$ is symmetric about $\alpha = 45^{\circ}$, i.e., $\sin(2\alpha) = \sin(2(90^{\circ}-\alpha))$.
  • The trajectory equation can be used to compute the angle given target coordinates (x, y) by solving the tangent form above.

Parabola fit from three trajectory points and extracting initial conditions

• Idea: A projectile path in the ideal model is a parabola: y = a x^2 + b x + c.

• If you have three points on the trajectory, you can determine a, b, c by solving the linear system:

  • Given points (x1, y1), (x2, y2), (x3, y3):

  egin{cases}
  y1 = a x1^2 + b x1 + c \ y2 = a x2^2 + b x2 + c \
  y3 = a x3^2 + b x_3 + c
  \end{cases}

• Once a, b, c are found, relate to the standard projectile form

  • From the path equation: $y = y<em>0 + (\tan \alpha) x - \frac{g x^2}{2 v</em>0^2 \cos^2\alpha}$
  • In the parabola form, the correspondence is:
  • Coefficient of x is $b = \tan\alpha$, so $\alpha = \arctan(b)$.
  • Coefficient of x^2 is $a = -\frac{g}{2 v<em>0^2 \cos^2\alpha} = -\frac{g}{2 v</em>0^2} (1 + \tan^2\alpha) = -\frac{g}{2 v_0^2} (1 + b^2)$
  • Constant term $c = y_0$ (launch height).

• From a and b, you can solve for the initial speed v0:

  • Using the relation for a:
    $a = - \frac{g}{2 v_0^2} (1 + b^2)$
  • Solve for v0^2:
    $v_0^2 = - \frac{g (1 + b^2)}{2 a}$

• Summary procedure:

  • Fit the trajectory data to the parabola y = a x^2 + b x + c using three chosen points.
  • Compute launch angle: $\alpha = \arctan(b)$.
  • Compute initial speed: $v_0^2 = - \frac{g (1 + b^2)}{2 a}$, provided a < 0 (a negative for an opening downward parabola).

• This approach connects data points to the underlying launch conditions without needing to measure angle directly at launch.

Connections to foundational principles

• The river-crossing examples illustrate decomposition of motion into perpendicular components (along and across the flow) and how to combine them via vector addition.
• The projectile motion section ties into fundamental kinematics: constant horizontal velocity, vertical acceleration due to gravity, and the resulting parabolic trajectory.
• The idea of avoiding manipulation of multiple system components in isolation (and instead adjusting interfaces or the system itself) echoes the engineering principle of designing robust, maintainable systems.

Practical takeaways and exam-ready points

• Black box vs white box mindset helps diagnose and fix system glitches by targeting interfaces or processes rather than individuals.
• In multi-system problems, separate into independent components (e.g., across width vs downstream drift) to simplify analysis.
• For projectile problems:
  • Always set up component velocities: $v<em>{0x} = v</em>0 \cos\alpha, \quad v<em>{0y} = v</em>0 \sin\alpha$.
  • Use the horizontal range formula and the trajectory equation to connect initial conditions to observed targets.
  • Remember the two-angle property for a given range (high vs low trajectory).
• When data points are available, parabola fitting can back out launch angle and speed using the relationships between the parabola coefficients and the physical parameters as shown above.

Quick reference formulas (summary)

• Velocity components:

$v<em>{0x} = v</em>0 \cos\alpha, \quad v<em>{0y} = v</em>0 \sin\alpha$

• Motion equations:
  x(t) = x0 + v{0x} t,

\quad y(t) = y0 + v{0y} t - \frac{1}{2} g t^2

• Horizontal range:
  $R = \frac{v_0^2}{g} \sin(2\alpha)$
  -\n- Trajectory (y as a function of x):

$y - y<em>0 = x \tan\alpha - \frac{g x^2}{2 v</em>0^2 \cos^2\alpha}$

• Parabolic form by substituting t = \tan\alpha:
  $a t^2 + b t + c = 0, \quad a = \frac{g x^2}{2 v<em>0^2}, \quad b = -x, \quad c = (y - y</em>0) + \frac{g x^2}{2 v_0^2}$
• Solve for tanα:
  $\tan\alpha = \frac{x \pm \sqrt{x^2 - 4ac}}{2a}$
• Parabola-fit to find initial speed from three points:
  • Fit: $y = a x^2 + b x + c$
  • Then $\alpha = \arctan(b)$ and
  • $v_0^2 = - \frac{g (1 + b^2)}{2 a}$
• Conceptual reminders:
  • Sine of twice the angle identity and the symmetry of high/low ballistic trajectories.
  • In multi-system problems, aim to modify the system or interfaces to avoid per-person workarounds and reduce the chance of human error.