Collision Course Teacher Guide Collision Course
Teacher Facilitation Guide
10th Grade Advanced Math
Linear-Quadratic Systems
Learning Objectives
Model real-world projectile motion using quadratic equations and linear flight paths.
Solve systems of linear-quadratic equations using the substitution method.
Critically evaluate mathematical solutions based on physical constraints (e.g., negative time).
Pacing & Procedure
05 min
Warm-up: The Visual System
Use Slide 3 to prompt students to visualize a ball (parabola) and a laser pointer (line). Ask: "Where do they meet? How many times can they meet?"
10 min
Video Refresher: Substitution Mechanics
Watch the provided video from 4:13 to 7:16. Focus on the algebraic steps of substitution and the application of the Quadratic Formula.
25 min
Activity: Target Practice
Students work through the word problem on the worksheet. They must set equations equal, rearrange into standard form, and solve for 't'.
05 min
Closure: The Physical Reality Check
Debrief the results. Focus on Scenario B (where one solution is negative). Why does the math give us a negative time, and why do we discard it?
Materials Checklist
Target Practice Slides
Collision Course Worksheet
System Solver Toolkit (Reference)
Scientific/Graphing Calculators
Key Misconceptions
"Always two answers": Students often think systems MUST have two solutions. Remind them of tangent (1) and non-intersecting (0) cases.
Negative Time: Students may try to force a negative solution into the context. Emphasize that \(t=0\) marks the "start" of our physical observation.
Critical Discussion Prompts
During the Video (Pause at 5:40):
"Before we plug into the formula, what do 'a', 'b', and 'c' represent in our projectile motion context? (Initial height, velocity, gravity etc.)"
The "Time Machine" Question:
"If we found \(t = -2\), does that mean the drone and ball hit 2 seconds before the throw? Mathematically, the paths would have met if they existed before \(t=0\)."
The Substitution Sequence
Step 1
Set Equal
\(h_1(t) = h_2(t)\)
Step 2
Zero Form
\(at^2 + bt + c = 0\)
Step 3
Quadratic Formula
\(t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
Step 4
Context Check
Is \(t > 0\)?
Target Practice Slides Collision Course
Linear-Quadratic Systems in Flight
Target Objectives
ALGEBRA
Master the substitution method for mixed systems.
PHYSICS
Interpret mathematical solutions as time and height.
"Math describes the universe, but logic determines its limits."
01 Visual Scenario
Imagine two paths in the air:
Path Alpha (Ball)
A high-arc parabolic lob thrown from 5 feet up.
Path Beta (Laser/Drone)
A constant linear ascent from 10 feet up.
Will they collide? Where?
Refining the Math
We've seen the visual intersection. Now, let's refresh our memory on the Substitution Method to find the exact coordinates of impact.
Substitution
Quadratic Formula
Algebraic Refresher
Focus Segment: 4:13 — 7:16
Pay attention to how the linear term is moved to standard form.
Embedded media
Mission: Target Practice
Worksheet Instructions
01
Set the drone and ball equations equal to find the time of impact.
02
Convert to Standard Form [\(ax^2+bx+c=0\)].
03
Solve for t and check the height h.
Watch out for "negative time" results. Are they physically real?
Math vs. Reality
The Mathematical Solution:
The intersection point on the coordinate plane exists wherever the two curves meet, including negative x-values.
The Physical Reality:
Time begins at \(t=0\). Any solution where \(t < 0\) is an "extraneous" result in physics—it describes where they *would* have met in the past.
Collision Course Worksheet Target Practice
Intercepting Flight Paths • Linear-Quadratic Systems
Student Name
Date
Scenario Alpha: The Intercept
HIGH ARC
A ball is thrown with an initial height of 5 feet and a velocity of 32 ft/s. Its height \(h\) at time \(t\) is modeled by: \(h(t) = -16t^2 + 32t + 5\). A drone flies a linear path modeled by: \(h(t) = 4t + 10\).
1. Set the equations equal & rewrite in Standard Form
2. Solve for \(t\) (Show your quadratic formula work)
Conclusion:
At what positive time(s) does the collision occur? Round to two decimal places.
Scenario Beta: The Low Pass
CRITICAL ANALYSIS
Suppose the drone flies a lower path: \(h(t) = 4t + 2\). The ball follows the same parabolic path: \(h(t) = -16t^2 + 32t + 5\).
1. Solve the system algebraically
The Mathematical Answers:
List both \(t\) values found above.
\(t_1 = \)
\(t_2 = \)
The Physical Reality:
Based on the context, how many collisions actually happen?
Final Reflection
In math, a negative solution for \(t\) exists because the equations are functions that extend infinitely. In physics, why do we call these "extraneous" or "impossible" solutions? What happened at \(t=0\)?
System Solver Toolkit System Solver
Mission Toolkit & Reference
The Workflow
01
Isolate & Equate
Ensure both equations are solved for the same variable (usually \(y\) or \(h\)). Set them equal: \(Linear = Quadratic\).
02
Standard Form
Move everything to one side so the equation is set to zero: \(ax^2 + bx + c = 0\).
03
Identify a, b, c
Carefully label your coefficients. Watch your signs!
04
Solve & Check
Plug into the Quadratic Formula. Find \(t\), then plug \(t\) back into the linear equation to find height.
Calculator Quick-Start
Finding Intersections
1 Press Y= and enter both equations.
2 Press GRAPH to see the paths.
3 Press 2nd + TRACE (CALC).
4 Select 5: Intersect.
5 Follow prompts: First curve? Second curve? Guess?
Formula Cheat Sheet
\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Physical Reality Matrix
Mathematical Solution Physical Interpretation \(t > 0\) Real collision. The objects meet after the experiment begins. \(t = 0\) Starting collision. The objects meet at the exact moment of launch. \(t < 0\) EXTRANEOUS. Describes an intersection that would have occurred before launch. Discard.
Collision Course Answer Key Collision Course
Official Answer Key • Teacher Resource
Unit Reference
Mixed Systems 2.0
A
Scenario Alpha
Step 1: Set Equal & Simplify
\(-16t^2 + 32t + 5 = 4t + 10\)
// Subtract 4t and 10 from both sides
\(-16t^2 + 28t - 5 = 0\)
\(16t^2 - 28t + 5 = 0\)
Step 2: Quadratic Formula
\(a=16, b=-28, c=5\)
\(t = \frac{28 \pm \sqrt{(-28)^2 - 4(16)(5)}}{2(16)}\)
\(t = \frac{28 \pm \sqrt{784 - 320}}{32} = \frac{28 \pm \sqrt{464}}{32}\)
\(t_1 \approx 0.20s \quad t_2 \approx 1.55s\)
Result: Two real, positive solutions. The ball hits the drone on the way up and again on the way down.
B
Scenario Beta
Step 1: Set Equal & Simplify
\(-16t^2 + 32t + 5 = 4t + 2\)
// Subtract 4t and 2 from both sides
\(-16t^2 + 28t + 3 = 0\)
\(16t^2 - 28t - 3 = 0\)
Step 2: Quadratic Formula
\(a=16, b=-28, c=-3\)
\(t = \frac{28 \pm \sqrt{784 - 4(16)(-3)}}{32}\)
\(t = \frac{28 \pm \sqrt{784 + 192}}{32} = \frac{28 \pm \sqrt{976}}{32}\)
\(t \approx 1.85s \quad t \approx -0.10s\)
MATH ANSWERS
\(t = 1.85, -0.10\)
PHYSICS ANSWER
Only 1 collision (\(t=1.85s\))
Final Reflection Key
The negative solution (\(t = -0.10\)) represents where the mathematical paths intersect if time existed before the start of the experiment. Physically, the ball was launched at \(t=0\), so any time less than zero is impossible. This solution is extraneous to our physical model.
