Motion Vector Slides Vector Motion
Defining Velocity and Acceleration in 3D Space
The Question of "Where?"
Imagine a drone flying through a canyon. To describe its path, a simple number isn't enough. We need a Vector-Valued Function.
\(\vec{r}(t) = \langle x(t), y(t), z(t) \rangle\)
How do we find out how fast it's changing position?
Velocity: The First Derivative
The **velocity vector** \(\vec{v}(t)\) is the instantaneous rate of change of the position vector \(\vec{r}(t)\).
\(\vec{v}(t) = \vec{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle\)
Direction: Tangent to the path of motion.
Magnitude: Represents the instantaneous speed.
Acceleration: The Second Derivative
The **acceleration vector** \(\vec{a}(t)\) measures how the velocity is changing—both in magnitude (speed) and direction.
\(\vec{a}(t) = \vec{v}'(t) = \vec{r}''(t)\)
*Note: \(\vec{a}(t)\) points toward the "inside" of the curve where the particle is turning.*
Acceleration Vector Visual
Example: Helical Path
Position: \(\vec{r}(t) = \langle \cos(t), \sin(t), t \rangle\)
Velocity
\(\vec{v}(t) = \langle -\sin(t), \cos(t), 1 \rangle\)
Acceleration
\(\vec{a}(t) = \langle -\cos(t), -\sin(t), 0 \rangle\)
Notice: The acceleration points directly back toward the \(z\)-axis!
Motion Vectors Worksheet MOTION VECTORS
LAB 01: VELOCITY & ACCELERATION DERIVATIVES
NAME: _________________________________
DATE: _________________________________
KINEMATIC DEFINITIONS
Position
\(\vec{r}(t)\)
Velocity
\(\vec{r}'(t)\)
Acceleration
\(\vec{r}''(t)\)
1
Particle Analysis: Polynomial Motion
A particle moves along a path described by \(\vec{r}(t) = \langle t^3 - 3t, 2t^2, \ln(t+1) \rangle\) for \(t \ge 0\).
A. Determine the velocity vector \(\vec{v}(t)\).
B. Determine the acceleration vector \(\vec{a}(t)\).
C. Find the acceleration vector at time \(t = 1\).
2
Geometric Interpretation
Consider a particle moving along the path \(\vec{r}(t) = \langle 4\cos(t), 4\sin(t), 3t \rangle\).
A. Show that the velocity vector and acceleration vector are always perpendicular to each other in the \(xy\)-plane (i.e., ignore the \(z\)-component for the dot product).
B. Sketch the path of the particle and draw the velocity and acceleration vectors at \(t = \pi/2\). Label your axes.
CRITICAL THINKING
If a particle is moving at a constant speed, does that mean its acceleration vector must be zero? Explain using a vector-valued function as an example.
Motion Vectors Key ANSWER KEY
LAB 01: VELOCITY & ACCELERATION DERIVATIVES
TEACHER RESOURCE
VECTOR MOTION MECHANICS
1
Particle Analysis: Polynomial Motion
A. Velocity Vector \(\vec{v}(t)\)
\(\vec{r}'(t) = \langle \frac{d}{dt}(t^3-3t), \frac{d}{dt}(2t^2), \frac{d}{dt}(\ln(t+1)) \rangle\)
\(\vec{v}(t) = \langle 3t^2 - 3, 4t, \frac{1}{t+1} \rangle\)
B. Acceleration Vector \(\vec{a}(t)\)
\(\vec{v}'(t) = \langle \frac{d}{dt}(3t^2-3), \frac{d}{dt}(4t), \frac{d}{dt}((t+1)^{-1}) \rangle\)
\(\vec{a}(t) = \langle 6t, 4, -\frac{1}{(t+1)^2} \rangle\)
C. Acceleration at \(t = 1\)
Substitute \(t = 1\) into \(\vec{a}(t)\):
\(\vec{a}(1) = \langle 6(1), 4, -\frac{1}{(1+1)^2} \rangle = \langle 6, 4, -0.25 \rangle\)
2
Geometric Interpretation
A. Dot Product Proof
\(\vec{v}_{xy}(t) = \langle -4\sin(t), 4\cos(t) \rangle\)
\(\vec{v} \cdot \vec{a} = (-4\sin t)(-4\cos t) + (4\cos t)(-4\sin t) = 16\sin t\cos t - 16\sin t\cos t = 0\)
Since the dot product is zero, the vectors are orthogonal (perpendicular).
B. Sketch Details
Path: Helix climbing the \(z\)-axis. At \(t = \pi/2\):
CRITICAL THINKING ANSWER
No. Acceleration is the rate of change of the velocity vector. Even if speed (the magnitude) is constant, if the direction of motion is changing, the velocity vector is changing, and thus acceleration is non-zero.
Example: Uniform circular motion \(\vec{r}(t) = \langle \cos t, \sin t \rangle\) has constant speed 1, but acceleration is \(\langle -\cos t, -\sin t \rangle\).
Distance and Speed Slides Path Analysis
Speed vs. Velocity and the Calculus of Distance
What is Speed?
While Velocity is a vector, Speed is a scalar. It is the magnitude of the velocity vector.
\(s(t) = \|\vec{v}(t)\|\)
Speed tells us how fast the odometer is ticking, regardless of which way the car is turning.
In 3D components:
\(s(t) = \sqrt{(x')^2 + (y')^2 + (z')^2}\)
Arc Length: Total Distance
To find the total distance traveled along a curved path from \(t=a\) to \(t=b\), we integrate the speed function.
\(L = \int_{a}^{b} \|\vec{v}(t)\| \, dt\)
Calculates path length (odometer)
Not the same as displacement!
Distance vs. Displacement
Displacement
The straight-line change in position.
\(\|\vec{r}(b) - \vec{r}(a)\|\)
Total Distance
The total path length traveled.
\(\int_{a}^{b} s(t) \, dt\)
"Displacement cares where you ended. Distance cares how you got there."
The Racing Drone Challenge
A drone's battery lasts for 500 meters of flight. If it follows the path \(\vec{r}(t) = \langle 10t, 2t^2, 0.5t^3 \rangle\), for how long can it stay in the air?
Step 1
Find Velocity \(\vec{v}(t)\)
Step 2
Find Speed \(\|\vec{v}(t)\|\)
Step 3
Integrate and solve for \(t\)
Path Length Worksheet PATH ANALYSIS
CALCULATING SPEED & TOTAL ARC LENGTH
UNIT: VECTOR CALCULUS
LAB: 02
ENGINEER: _________________________________
DATE: _________________________________
Phase 1: Instantaneous Speed
Find the speed function \(s(t) = \|\vec{v}(t)\|\) for a particle with position vector \(\vec{r}(t) = \langle 3t, 4\sin(t), 4\cos(t) \rangle\).
Show derivation of velocity:
Calculate magnitude (simplify!):
FINAL SPEED \(s(t)\) =
Phase 2: Arc Length
Calculate the total distance traveled by the particle from Part 1 over the time interval \(0 \le t \le 2\pi\).
// Setup the integral for Arc Length (L):
// Evaluate the integral:
DISPLACEMENT CHECK
Using the same function \(\vec{r}(t) = \langle 3t, 4\sin(t), 4\cos(t) \rangle\), calculate the displacement (straight-line distance) from \(t=0\) to \(t=2\pi\). How does it compare to the total distance?
Position at \(t=0\)
Position at \(t=2\pi\)
Displacement Magnitude
Conclusion: Why is the displacement value different from the arc length value?
Path Length Key Analysis Key
LAB 02: SPEED & ARC LENGTH SOLUTIONS
TEACHER COPY
PHASE 1: INSTANTANEOUS SPEED
Velocity Derivation
\(\vec{r}(t) = \langle 3t, 4\sin t, 4\cos t \rangle\)
Speed Magnitude
\(s(t) = \sqrt{3^2 + (4\cos t)^2 + (-4\sin t)^2}\)
\(s(t) = 5\)
Constant Speed Identified
PHASE 2: ARC LENGTH EVALUATION
Integral Setup & Calculation
\(L = \int_{0}^{2\pi} 5 \, dt\)
\(L = 10\pi \approx 31.42 \text{ units}\)
PHASE 3: DISPLACEMENT COMPARISON
Position Values
\(\vec{r}(0) = \langle 0, 0, 4 \rangle\)
Vector Change
\(\Delta \vec{r} = \langle 6\pi, 0, 0 \rangle\)
CONCLUSION FOR STUDENTS:
The arc length (\(10\pi\)) is longer than the displacement (\(6\pi\)) because the particle is spiraling (helical motion). The displacement only measures the straight distance along the axis of the spiral, while the arc length counts the "winding" travel around the cylinder.
Projectiles in Space Slides Target Acquired
3D Projectile Modeling with Vector-Valued Functions
The Physics of Freefall
In a vacuum, the only force acting on a projectile is gravity, acting in the negative \(z\) direction (or \(y\) in 2D).
// Acceleration Vector
\(\vec{a}(t) = \langle 0, 0, -g \rangle\)
where \(g \approx 9.8 \, m/s^2\) or \(32 \, ft/s^2\)
Integration Steps:
1
Integrate \(\vec{a}(t)\) to find \(\vec{v}(t)\)
2
Apply initial velocity \(\vec{v}_0\)
3
Integrate \(\vec{v}(t)\) to find \(\vec{r}(t)\)
The Kinematic Vector Equation
Master Formula
\(\vec{r}(t) = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{g} t^2\)
Initial Position
\(\vec{r}_0 = \langle x_0, y_0, z_0 \rangle\)
Initial Velocity
\(\vec{v}_0 = \langle v_{0x}, v_{0y}, v_{0z} \rangle\)
Gravity Effect
\(\frac{1}{2} \langle 0, 0, -g \rangle t^2\)
Launching at Angles
In 3D, initial velocity depends on an elevation angle (\(\phi\)) and an azimuth angle (\(\theta\)).
\(v_{0x} = \|\vec{v}_0\| \cos\phi \cos\theta\)
Think of it like spherical coordinates for velocity!
Solving the System
Max Height
Find \(t\) when the vertical component of velocity is zero.
\(v_{0z} - gt = 0\)
Flight Time
Find \(t\) when the vertical component of position returns to zero.
\(z(t) = 0\)
Range
Calculate horizontal distance at the end of flight time.
\(d = \sqrt{x(t_{end})^2 + y(t_{end})^2}\)
Ballistics Lab Worksheet BALLISTICS LAB
MODELING 3D PROJECTILE FLIGHT
IDENT: ______________________
STATION: ____________________
SPECS
A virtual projectile is launched from position \(\vec{r}_0 = \langle 0, 0, 10 \rangle\) meters at \(t=0\). The initial velocity vector is \(\vec{v}_0 = \langle 20, 15, 49 \rangle\) m/s. Use \(g = 9.8 \, m/s^2\).
1. Derive the position vector function \(\vec{r}(t)\)
2. Find the velocity vector function \(\vec{v}(t)\)
MISSION DATA ANALYSIS
A. Time of Impact
Set \(z(t) = 0\) and solve for \(t > 0\). Use the quadratic formula if necessary.
B. Impact Velocity
Calculate \(\vec{v}(t_{impact})\). Then find the impact speed \(\|\vec{v}\|\).
C. Maximum Height reached
Find \(t\) when \(v_z(t) = 0\), then find \(z(t)\).
D. Total Horizontal Range
Distance from \((0,0)\) to impact \((x_{end}, y_{end})\).
TACTICAL ADJUSTMENT
A mountain peak is located at \((200, 150, 100)\). Using your position function \(\vec{r}(t)\), determine if the projectile passes safely over the mountain or if it crashes into it. Show your mathematical evidence.
Ballistics Lab Key Ballistics Key
MISSION DATA VALIDATION: PROJECTILE 3D
OFFICIAL RECORD
1. Derivation of Functions
Position \(\vec{r}(t)\)
\(\vec{r}(t) = \langle x_0 + v_{0x}t, y_0 + v_{0y}t, z_0 + v_{0z}t - \frac{1}{2}gt^2 \rangle\)
Velocity \(\vec{v}(t)\)
\(\vec{v}(t) = \vec{r}'(t)\)
2. Flight Analysis
A. TIME OF IMPACT
\(10 + 49t - 4.9t^2 = 0\)
C. MAX HEIGHT
\(49 - 9.8t = 0 \Rightarrow t = 5 \text{ s}\)
B. IMPACT VELOCITY
\(\vec{v}(10.2) = \langle 20, 15, -50.96 \rangle\)
D. TOTAL RANGE
\(x(10.2) = 204, y(10.2) = 153\)
Tactical Resolution
Mountain located at x=200, y=150.
When is the projectile at these coordinates?
\(20t = 200 \Rightarrow t = 10\text{s}\).
Calculate height at t=10:
RESULT: CRASH. The projectile is at 10m height when the mountain peak is at 100m. It hits the base/side of the mountain.
Turning Forces Math Slides The Anatomy of G-Force
Tangential and Normal Components of Acceleration
Two Reasons to Accelerate
1. Changing Speed
The "push back in your seat" feeling. This is the Tangential Component (\(a_T\)).
2. Changing Direction
The "slammed against the door" feeling during a turn. This is the Normal Component (\(a_N\)).
Calculus of \(a_T\)
The tangential component is simply the derivative of speed with respect to time. It acts in the direction of motion.
\(a_T = \frac{ds}{dt}\)
Vector Method:
\(a_T = \frac{\vec{v} \cdot \vec{a}}{\|\vec{v}\|}\)
This uses the projection of acceleration onto velocity.
Calculus of \(a_N\)
The normal component is the part of acceleration perpendicular to motion. It relates to curvature (\(\kappa\)).
\(a_N = \sqrt{\|\vec{a}\|^2 - a_T^2}\)
Vector Method:
\(a_N = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|}\)
*Only applicable in 3D using cross product.
The Total Vector Sum
\(\vec{a} = a_T \vec{T} + a_N \vec{N}\)
\(\vec{T}\) Unit Tangent
Points where you are going.
\(\vec{N}\) Principal Normal
Points where you are turning.
Turning Forces Worksheet Vector Decomposition
LAB 04: TANGENTIAL & NORMAL ACCELERATION
IDENT: ______________________
DATE: _______________________
Tangential \(a_T = \frac{\vec{v} \cdot \vec{a}}{\|\vec{v}\|}\)
Normal \(a_N = \sqrt{\|\vec{a}\|^2 - a_T^2}\)
1
Circular Path Analysis
A particle moves according to \(\vec{r}(t) = \langle 3\cos(t^2), 3\sin(t^2) \rangle\). Note the \(t^2\) argument!
A. Find \(\vec{v}(t)\) and \(\vec{a}(t)\)
B. Find the speed \(s(t)\)
C. Calculate \(a_T\). Is the particle speeding up, slowing down, or moving at constant speed?
D. Calculate \(a_N\). How does the turning force change over time?
2
Geometric Breakdown
At a specific moment \(t=t_0\), a particle has velocity \(\vec{v} = \langle 3, 4, 0 \rangle\) and acceleration \(\vec{a} = \langle 1, 2, 5 \rangle\).
A. Find the tangential component of acceleration.
B. Find the normal component of acceleration.
Decomposition Verification
Verify that \(\vec{a} = a_T \vec{T} + a_N \vec{N}\) by calculating \(\|\vec{a}\|^2\) and comparing it to \(a_T^2 + a_N^2\).
ENGINEERING INSIGHT
If a particle is moving along a straight line, what must be true about the normal component of acceleration? Explain your answer using the cross product formula \(a_N = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|}\).
Turning Forces Key Decomposition Key
LAB 04: TANGENTIAL & NORMAL ACCELERATION SOLUTIONS
MASTER FILE
1. Circular Path Analysis
Vectors
\(\vec{r}(t) = \langle 3\cos(t^2), 3\sin(t^2) \rangle\)
Speed
\(s(t) = \sqrt{(-6t\sin(t^2))^2 + (6t\cos(t^2))^2}\)
Tangential \(a_T\)
\(a_T = \frac{d}{dt}(6t) = \mathbf{6}\)
Constant speeding up.
Normal \(a_N\)
\(\|\vec{a}\|^2 = (-6\sin u - 12t^2\cos u)^2 + (6\cos u - 12t^2\sin u)^2\)
2. Moment Analysis
Step A: Tangential
\(\vec{v} \cdot \vec{a} = (3)(1) + (4)(2) + (0)(5) = 3 + 8 + 0 = 11\)
Step B: Normal
\(\|\vec{a}\|^2 = 1^2 + 2^2 + 5^2 = 30\)
Insight Answer Key
If a particle moves in a straight line, its velocity \(\vec{v}\) and acceleration \(\vec{a}\) must be parallel (collinear). The cross product of two parallel vectors is always the zero vector. Therefore, \(\|\vec{v} \times \vec{a}\| = 0\), which results in \(a_N = 0\). This makes physical sense because \(a_N\) represents the force of turning, and there is no turning in a straight line.
Orbits and Gravity Slides Celestial Mechanics
Orbits, Gravity, and Circular Vector Dynamics
The Geometry of Orbits
For a stable circular orbit, the acceleration vector points directly toward the center of the mass being orbited.
// Vector Form
\(\vec{r}(t) = R \langle \cos(\omega t), \sin(\omega t) \rangle\)
Centripetal Force
Speed-Curvature Relationship
The normal component of acceleration (\(a_N\)) in a circular orbit is directly related to the orbital speed \(v\) and the radius \(R\).
\(a_N = \frac{v^2}{R}\)
Low Orbit
Small \(R\) requires very high speed \(v\) to maintain \(a_N\).
High Orbit
Large \(R\) allows for lower speed \(v\) (Geostationary).
Beyond Circles: Ellipses
Most orbits are elliptical. In an ellipse, the acceleration vector still points toward the focus (the Sun/Earth), but it is no longer perpendicular to the velocity vector.
This means speed is changing throughout the orbit! Faster at periapsis, slower at apoapsis.
Kepler's Second Law: Areas and Speeds
Satellite Telemetry
If a satellite's acceleration vector \(\vec{a}\) is exactly \(-k \vec{r}\), what does that tell you about its path?
It is a perfect circular orbit.
Speed is constant (\(a_T = 0\)).
The orbit is stable indefinitely.
Satellite Tracking Worksheet Satellite Tracking
ORBITAL DYNAMICS & VECTOR STABILITY
ORBITAL REGION: LEO-244
SATELLITE ID: __________________
TRACKING DATE: _______________
Telemetry Feed: Circular Orbit
A communication satellite follows a path given by \(\vec{r}(t) = \langle 400\cos(0.02t), 400\sin(0.02t) \rangle\), where distance is in kilometers and time is in minutes.
1. Determine the Orbital Speed \(s(t)\)
2. Find the Acceleration Vector \(\vec{a}(t)\)
Vector Relationship Analysis
Show that \(\vec{a}(t)\) is proportional to \(\vec{r}(t)\). Specifically, find the constant \(k\) such that \(\vec{a}(t) = -k\vec{r}(t)\). What is the physical significance of the negative sign?
Case Study: Elliptical Drift
Another probe is tracking an elliptical path \(\vec{r}(t) = \langle 500\cos(0.01t), 300\sin(0.01t) \rangle\).
A. SPEED VARIATION
Calculate the speed at \(t=0\) and \(t=50\pi\). Does the satellite maintain a constant speed?
B. ACCELERATION COMPONENTS
At \(t = 25\pi\), calculate the tangential acceleration \(a_T\). Explain what this means for the probe's mission.
Keplerian Logic
If the Earth's gravity were to suddenly weaken, the normal component of acceleration \(a_N\) would decrease. Based on the formula \(a_N = v^2/R\), what would happen to the satellite's orbital radius if it tried to maintain its current speed?
Satellite Tracking Key Tracking Log Key
ORBITAL DYNAMICS VALIDATION SYSTEM
ACCESS: GRANTED
1. Circular Orbit Analysis
Speed Calculation
\(\vec{v}(t) = \langle -8\sin(0.02t), 8\cos(0.02t) \rangle\)
Acceleration & Proportion
\(\vec{a}(t) = \langle -0.16\cos(0.02t), -0.16\sin(0.02t) \rangle\)
Physical Significance
The negative sign indicates the acceleration vector is in the opposite direction of the position vector, meaning it points directly toward the center (Earth), acting as a centripetal force.
2. Elliptical Drift Solutions
A. SPEED VARIATION
\(\vec{v}(t) = \langle -5\sin(0.01t), 3\cos(0.01t) \rangle\)
B. TANGENTIAL ACCEL
At \(t = 25\pi\): \(0.01t = \pi/4\).
Reflective Analysis Answer
From \(a_N = v^2/R\), we see that \(R = v^2/a_N\). If \(a_N\) (the pull of gravity) decreases while \(v\) stays constant, the denominator decreases, causing the radius \(R\) to increase. The satellite would spiral outward into a higher orbit or escape entirely if the force is no longer sufficient to maintain a curve.
