Speed Shift Science • Lesson • Lenny.com
Speed Shift Science A comprehensive physics lesson focused on understanding acceleration as the rate of change of velocity, calculating acceleration using kinematic formulas, and determining acceleration from the slope of velocity-time graphs.
CM Chuguna Murugesan
Acceleration Fundamentals Handout
A comprehensive two-page conceptual guide and reference sheet explaining what acceleration is in depth, the physics of velocity change, intuitive sensation of acceleration, unit analysis, the algebraic formula, and velocity-time graph slope calculations.
Acceleration Fundamentals Teacher Guide
A two-page teacher instructional and concept facilitation guide covering deep pedagogy of acceleration, Socratic questioning, unit derivations, velocity-time graph slope analysis, and common misconception breakdowns.
Acceleration Practice Worksheet
A four-page comprehensive student practice assignment containing 6 word problems and 5 graphical analysis problems across three velocity-time graphs, with ample handwriting space and no formulas or hints provided.
Acceleration Practice Solutions
A four-page comprehensive teacher solution and scoring guide matching the four-page student assignment, with complete algebraic solutions, rubric criteria, and common errors for all 11 problems.
Speed Shift Science A comprehensive physics lesson focused on understanding acceleration as the rate of change of velocity, calculating acceleration using kinematic formulas, and determining acceleration from the slope of velocity-time graphs.
CM Chuguna Murugesan
Acceleration Fundamentals Handout Physics Kinematics • Core Conceptual & Reference Guide Page 1 of 2
Understanding Acceleration: The Complete Guide
What acceleration really means, how your body senses it, units, and formulas.
Name: __________________
Date: ____________
Period: _______
1
What is Acceleration? (The Physical Definition)
In physics, acceleration (\(a\)) is the rate of change of velocity over time. Velocity is a vector consisting of both speed and direction . An object accelerates whenever its speed changes, its direction changes, or both change:
1. Speeding Up
Velocity and acceleration point in the same direction . Speed increases.
Example: Green light sprint
2. Slowing Down
Velocity and acceleration point in opposite directions . Speed decreases.
Example: Braking for a red light
3. Changing Direction
Even at constant speed , turning changes velocity direction, which requires acceleration!
Example: Roundabout turn
The Body as an Acceleration Detector: You cannot feel velocity! On a jet cruising smoothly at \(550\text{ mph}\), a cup of juice sits still because \(a = 0\text{ m/s}^2\). You only feel a physical push when the plane accelerates: pushed into your seat during takeoff, thrown forward during braking, or tilted during a banking turn.
2
Demystifying the Units: What Does \(\text{m/s}^2\) Mean?
Unit: \(\text{m/s}^2 = \frac{\text{m/s}}{\text{s}}\)
Meters per second squared means "meters per second, each second" . It specifies how much speed changes every second.
\(+5\text{ m/s}^2\) means gaining \(5\text{ m/s}\) every single second.
Second-by-Second Velocity Progression (\(a = +5\text{ m/s}^2\))
Start (\(t=0\)) 0 m/s
After 1s +5 m/s
After 2s +10 m/s
After 3s +15 m/s
3
The Acceleration Formula & Sign Conventions
Algebraic Equation
\(a = \dfrac{\Delta v}{\Delta t} = \dfrac{v_f - v_i}{t}\)
Final Velocity minus Initial Velocity
\(a\): Acceleration (\(\text{m/s}^2\))
\(t\): Time interval (\(\text{s}\))
\(v_i\): Initial velocity (\(\text{m/s}\))
\(v_f\): Final velocity (\(\text{m/s}\))
Positive Acceleration (\(a > 0\)): Speeding up in positive direction (\(v_f > v_i\)) or slowing down in reverse direction.
Negative Acceleration (\(a < 0\)): Slowing down in positive direction (\(v_f < v_i\)) or speeding up in reverse direction.
4
Worked Example: Step-by-Step Calculation & Meaning
Model Walkthrough
Problem: A train traveling forward at \(30\text{ m/s}\) applies brakes and smoothly halts to a stop (\(0\text{ m/s}\)) in \(5.0\text{ seconds}\). Calculate and interpret its acceleration.
1. List Givens
\(v_i = 30\text{ m/s}\)
2. State Formula
\(a = \dfrac{v_f - v_i}{t}\)
3. Substitute & Solve
\(a = \dfrac{0 - 30}{5.0} = \dfrac{-30}{5.0}\)\(a = -6.0\text{ m/s}^2\)
4. Physical Meaning
The train loses \(6.0\text{ m/s}\) of forward speed every second until stopped.
Turn to Page 2 for Velocity-Time Graphs, Slope Analysis, and Comprehensive Visual Reference →
Part 2: Graphical Analysis & Slope Determination Page 2 of 2
Finding Acceleration from Velocity-Time Graphs
The slope of a velocity-time line corresponds directly to physical acceleration.
Kinematics Reference Guide
5
Why Does Slope Equal Acceleration?
Definition of Slope
\(\text{Slope} = \dfrac{\text{Rise}}{\text{Run}} = \dfrac{\Delta v}{\Delta t} = \dfrac{v_2 - v_1}{t_2 - t_1} = \mathbf{a}\)
Rise = Velocity change • Run = Time elapsed
↗ Upward Slope \(a > 0\) Speeding up forward
→ Horizontal Line \(a = 0\text{ m/s}^2\) Constant velocity
↘ Downward Slope \(a < 0\) Slowing down
6
Complete Motion Journey: Drone Mission Velocity-Time Graph
Visual Demonstration
0 10 20 30 v (m/s) 0 2 4 6 8 Time (s) A B C
Segment A: Upward Slope (Speeding Up)
\(\text{Rise } \Delta v = 20 - 0 = +20\text{ m/s}\), \(\text{Run } \Delta t = 2 - 0 = 2\text{ s}\)
Segment B: Flat Horizontal Line (Cruising)
\(\text{Rise } \Delta v = 20 - 20 = 0\text{ m/s}\), \(\text{Run } \Delta t = 5 - 2 = 3\text{ s}\)
Segment C: Downward Slope (Decelerating)
\(\text{Rise } \Delta v = 0 - 20 = -20\text{ m/s}\), \(\text{Run } \Delta t = 8 - 5 = 3\text{ s}\)
7
Mastery Comparison: Position-Time vs. Velocity-Time Graphs
Cheat Sheet
Graph Visual Feature Position-Time (\(x\)-\(t\)) Meaning Velocity-Time (\(v\)-\(t\)) Meaning — Horizontal Line Object is stationary (at rest) Constant velocity (\(a = 0\text{ m/s}^2\)) / Straight Diagonal Line Constant velocity (steady speed) Constant acceleration (\(a = \text{const}\)) ◜ Curved Upward Line Accelerating (speed is increasing) Changing acceleration (jerk) Slope Calculation \(\text{Slope} = \dfrac{\Delta x}{\Delta t} = \text{Velocity } (v)\) \(\text{Slope} = \dfrac{\Delta v}{\Delta t} = \text{Acceleration } (a)\)
Finding Acceleration: \(a = \dfrac{v_f - v_i}{t}\)
Finding Final Velocity: \(v_f = v_i + (a \cdot t)\)
Finding Time Elapsed: \(t = \dfrac{v_f - v_i}{a}\)
Key Takeaway: Acceleration measures how rapidly velocity transforms. On any \(v\)-\(t\) graph, steepness equals acceleration! ✓ Reference Complete
Acceleration Fundamentals Teacher Guide Teacher Resource • Instructional & Conceptual Guide Page 1 of 2
Teaching Acceleration: Pedagogy & Insights
Instructional strategies, core physical intuitions, and misconception breakdowns.
Physics Kinematics
1
Core Teaching Strategy: Establishing Physical Intuition
Students routinely conflate velocity (how fast and what way you are moving) with acceleration (how rapidly your motion is altering). Use tactile, physical analogies before introducing equations:
1. The "Closed Airplane" Thought Experiment:
Ask students: "If you close your eyes on a smoothly cruising 550 mph jet, can you tell you are moving?" The answer is no. Constant velocity has zero acceleration (\(a = 0\)). Human inner-ear fluid and skin sensors only register net forces, which occur exclusively during acceleration!
2. The "Three Pedals" Vehicle Analogy:
A standard car has three acceleration controls: (1) The gas pedal (increases speed), (2) The brake pedal (decreases speed), and (3) The steering wheel (changes direction). All three produce physical acceleration.
2
Diagnostic Socratic Prompts & Misconceptions
Misconception: "Negative acceleration always means slowing down." Correction: Acceleration is a directional vector. If an elevator is moving downward (negative velocity) and speeds up downward, its acceleration is negative! A negative sign simply points in the chosen negative coordinate direction.
Misconception: "If velocity is zero, acceleration must be zero." Correction: When a ball is thrown straight upward, at the very peak its instantaneous velocity is \(0\text{ m/s}\), but gravity continues to accelerate it downward at \(9.8\text{ m/s}^2\). If \(a\) were zero at the peak, the ball would hover permanently!
3
Worked Example 1 Solution & Algebraic Nuance
Model Solution
Train Braking Problem Analysis: \(v_i = +30\text{ m/s}\), \(v_f = 0\text{ m/s}\), \(t = 5.0\text{ s}\)
Grading & Review Criteria:
Negative sign must be preserved (\(0 - 30\)).
Units must be \(\text{m/s}^2\), not \(\text{m/s}\).
Meaning: decelerating at \(6.0\text{ m/s}\) each sec.
Acceleration Practice Worksheet Physics Assignment • Kinematics Problem Set Page 1 of 4
Acceleration Practice Assignment
Solve each problem completely. State all givens, write your formula, and show your calculations with units.
Name: ___________________
Date: _________
Period: _____
1 Cheetah Sprint: Acceleration from Rest [ 3 Points ]
A cheetah stalks its prey and accelerates forward in a straight line from rest (\(0\text{ m/s}\)) to a top running velocity of \(25.5\text{ m/s}\) in exactly \(3.0\text{ seconds}\). Calculate the cheetah's acceleration.
\(v_i\): ______________
\(v_f\): ______________
\(t\): ______________
Show full mathematical calculation with units:
Final Answer: \(a =\) __________________________
2 Braking Vehicle: Negative Acceleration [ 3 Points ]
A sports sedan traveling forward at \(28.0\text{ m/s}\) applies its brakes to avoid road debris, slowing down uniformly to \(4.0\text{ m/s}\) over a duration of \(4.0\text{ seconds}\). (a) Calculate the vehicle's acceleration. (b) Explain why the answer is negative.
\(v_i\): ______________
\(v_f\): ______________
\(t\): ______________
Show full mathematical calculation with units:
Final Answer: \(a =\) ____________________ Explanation of sign: _____________________________________
3 Skateboarder Ramp: Finding Final Velocity [ 3 Points ]
A skateboarder pushes off down a ramp with an initial velocity of \(2.0\text{ m/s}\) and accelerates steadily at a rate of \(1.8\text{ m/s}^2\) for \(5.0\text{ seconds}\). What is the skateboarder's final velocity at the bottom of the ramp?
\(v_i\): ______________
\(a\): ______________
\(t\): ______________
Show full mathematical calculation with units:
Final Velocity: \(v_f =\) __________________________
Turn to Page 2 for Word Problems 4, 5, and 6 →
Kinematics Problem Set • Continued Page 2 of 4
Kinematics Word Problems (4 – 6)
Apply algebraic rearrangements to solve for time and initial velocity.
Name: ______________________
4 Runway Landing: Solving for Time [ 3 Points ]
A cargo aircraft touches down on a runway with an initial landing speed of \(70.0\text{ m/s}\). Its thrust reversers and wheel brakes produce a uniform acceleration of \(-3.5\text{ m/s}^2\) until the plane halts completely to rest (\(0\text{ m/s}\)). How many seconds does it take for the plane to stop?
Acceleration Practice Solutions Teacher Solutions & Scoring Rubric Page 1 of 4
Solutions: Word Problems (1 – 3)
Model responses, mathematical working, and scoring rubrics (9 points on Page 1).
Total: 35 Points Assignment
Problem 1: Cheetah Sprint Acceleration [ 3 Points ]
Givens & Formula: \(v_i = 0\text{ m/s}\), \(v_f = 25.5\text{ m/s}\), \(t = 3.0\text{ s}\)
Scoring Breakdown:
Listing givens correctly [1 pt]
Formula & substitution [1 pt]
Value \(8.5\) & unit \(\text{m/s}^2\) [1 pt]
Problem 2: Braking Sports Sedan (Deceleration) [ 3 Points ]
Givens & Formula: \(v_i = 28.0\text{ m/s}\), \(v_f = 4.0\text{ m/s}\), \(t = 4.0\text{ s}\)
Scoring Breakdown:
Listing givens correctly [0.5 pt]
Calculation & negative sign [1.5 pts]
Explanation of negative sign [1 pt]
Problem 3: Skateboarder Final Velocity [ 3 Points ]
Givens & Formula: \(v_i = 2.0\text{ m/s}\), \(a = 1.8\text{ m/s}^2\), \(t = 5.0\text{ s}\)
Scoring Breakdown:
Listing givens correctly [1 pt]
Correct formula isolation [1 pt]
Value \(11.0\) and unit \(\text{m/s}\) [1 pt]
Turn to Page 2 for Solutions to Word Problems 4, 5, and 6 →
Teacher Solutions & Scoring Rubric Page 2 of 4
Solutions: Word Problems (4 – 6)
Model responses for time and initial velocity derivations (9 points on Page 2).
9 Points Available
Problem 4: Cargo Plane Landing Time [ 3 Points ]
Givens & Formula: \(v_i = 70.0\text{ m/s}\), \(v_f = 0\text{ m/s}\), \(a = -3.5\text{ m/s}^2\)
Scoring Breakdown:
Turn to Page 2 for Velocity-Time Graph Instruction, Slope Analysis, and Formative Questions →
Part 2: Graph Pedagogy & Advanced Analysis Page 2 of 2
Graph Instruction & Slope Determination Guiding students to extract acceleration and displacement from motion graphs.
Drone Graph Master Solution & Coordinate Key Segment A (\(0\text{ s}\) to \(2\text{ s}\)): Coordinates: \((0, 0)\) to \((2, 20)\)
\(a_A = \dfrac{20 - 0}{2 - 0} = \mathbf{+10\text{ m/s}^2}\)
Uniform positive acceleration
Segment B (\(2\text{ s}\) to \(5\text{ s}\)): Coordinates: \((2, 20)\) to \((5, 20)\)
\(a_B = \dfrac{20 - 20}{5 - 2} = \mathbf{0\text{ m/s}^2}\)
Constant cruising velocity
Segment C (\(5\text{ s}\) to \(8\text{ s}\)): Coordinates: \((5, 20)\) to \((8, 0)\)
\(a_C = \dfrac{0 - 20}{8 - 5} = \mathbf{-6.67\text{ m/s}^2}\)
Uniform braking to full stop
Crucial Graph Traps & Student Hurdles Trap 1: The "Hills and Valleys" Illusion
Students see an upward slope on a \(v\)-\(t\) graph and imagine the drone is physically flying up a steep mountain hill. Remind them: the vertical axis is speed , not elevation or altitude! A car speeding up on flat highway has the exact same upward slope.
Trap 2: The Downward Slope vs. Moving Backwards
Students often assume a downward sloping line means traveling backwards. Clarify: as long as the line is above the horizontal time axis (\(v > 0\)), the object is still moving forward—it is just slowing down! Only if the line crosses below the time axis does reverse motion begin.
Extension Opportunity: Area Under the Curve (Displacement) Point out to advanced learners that while the slope yields acceleration, the geometric area bounded between the curve and the time axis equals the total distance (displacement) traveled:
Triangle A (0 to 2s): \(\frac{1}{2} \times 2 \times 20 = \mathbf{20\text{ m}}\)
Rectangle B (2 to 5s): \(3 \times 20 = \mathbf{60\text{ m}}\)
Triangle C (5 to 8s): \(\frac{1}{2} \times 3 \times 20 = \mathbf{30\text{ m}}\)
Total Mission Distance = \(20\text{ m} + 60\text{ m} + 30\text{ m} = \mathbf{110\text{ meters}}\)
Teaching Summary: Connect kinesthetic experience (feeling pushed) → verbal definition (rate of change) → math slope. ✓ Lesson Guide Ready
Show full mathematical calculation with units:
Total Time: \(t =\) __________________________
5 Motorcycle Surge: Finding Initial Velocity [ 3 Points ]
After steadily accelerating at \(2.5\text{ m/s}^2\) for a time interval of \(6.0\text{ seconds}\), a motorcycle reaches a cruising speed of \(32.0\text{ m/s}\). What was the motorcycle's initial velocity right before this acceleration began?
Show full mathematical calculation with units:
Initial Velocity: \(v_i =\) __________________________
6 Subway Departure: Acceleration & Rate of Gain [ 3 Points ]
An electric subway train leaves a terminal platform starting from rest (\(0\text{ m/s}\)) and accelerates along a straight track at \(1.5\text{ m/s}^2\) for \(12.0\text{ seconds}\).
Show full mathematical calculation with units:
(a) Final Velocity: \(v_f =\) ____________________ (b) Velocity gained per second: ______________
Turn to Page 3 for Graph Analysis Problems 7 through 10 →
Part 2: Graphical Analysis & Slope Page 3 of 4
Acceleration Practice: Graph Problems (7 – 10) Determine physical acceleration directly from the slope of each velocity-time graph segment.
Name: ______________________
Graph 1: Hyperloop Capsule Test Run Questions 7 & 8 [ 6 pts ]
0 15 30 45 v (m/s) 0 4 10 16 Time (s) Seg A Seg B Seg C
Question 7: Calculate Acceleration in Segment A (\(0\) to \(4\text{ s}\)) [ 3 pts ]
Read the graph to find the acceleration of the capsule during Segment A. Show all work with units:
Answer: \(a_A =\) ________________________
Question 8: Analyze Segment B (\(4\) to \(10\text{ s}\)) [ 3 pts ]
What is the acceleration during Segment B? Explain your reasoning based on the graph line:
Answer: \(a_B =\) _________ • Reason: ________________________
Graph 2: Emergency Braking & Reversal Questions 9 & 10 [ 6 pts ]
+20 +10 0 -10 v (m/s) 0 2 4 6 Time (s) Braking Reverse
Question 9: Calculate Slope of Braking Phase (\(t = 0\) to \(4\text{ s}\)) [ 3 pts ]
Read the graph to determine the acceleration during the braking phase from \(t = 0\text{ s}\) to \(t = 4\text{ s}\):
Answer: \(a =\) ________________________
Question 10: Interpreting the Zero Crossing [ 3 pts ]
(a) At what time does the vehicle stop? (b) What happens physically when \(v < 0\) (\(t > 4\text{ s}\))?
(a) \(t =\) _________ • (b) Meaning: ________________________
Turn to Page 4 for Multi-Stage Flight Graph Problem 11 →
Part 2: Graphical Analysis • Multi-Stage Flight Page 4 of 4
Graph Problem 11: Multi-Stage Drone Flight Analyze complex motion phases, determine peak accelerations, and compare slopes.
Name: ______________________
Graph 3: Autonomous Cargo Drone Velocity Profile Problem 11 [ 5 Points Total ]
0 8 16 24 v (m/s) 0 4 8 12 16 Time (s) Phase 1 Phase 2 Phase 3 Phase 4
(a) Steepest Descent: Calculate Acceleration in Phase 3 (\(t = 10\text{ s}\) to \(t = 12\text{ s}\)) [ 2 pts ]
Read the coordinates from the graph and calculate the acceleration during the steep braking phase. Show all work with units:
Phase 3 Acceleration: \(a_3 =\) __________________________
(b) Final Touchdown: Calculate Acceleration in Phase 4 (\(t = 12\text{ s}\) to \(t = 16\text{ s}\)) [ 2 pts ]
Read the coordinates from the graph and calculate the acceleration as the drone glides gently to rest:
Phase 4 Acceleration: \(a_4 =\) __________________________
(c) Comparative Slope & Physical Meaning [ 1 pt ]
Why is the acceleration magnitude in Phase 3 strictly greater than in Phase 4? Explain using the steepness of the line:
Explanation: __________________________________________________________________________________
Assignment Complete: 11 total problems (6 word problems + 5 graph analysis problems). Total score: 35 points. ✓ All 4 Pages Finished
Negative sign on \(a\) and \(\Delta v\) [1 pt]
Division producing positive time [1 pt]
Final value \(20.0\text{ s}\) with unit [1 pt]
Problem 5: Motorcycle Initial Velocity [ 3 Points ]
Givens & Formula: \(v_f = 32.0\text{ m/s}\), \(a = 2.5\text{ m/s}^2\), \(t = 6.0\text{ s}\)
Listing givens correctly [1 pt]
Algebraic formula setup [1 pt]
Value \(17.0\) and unit \(\text{m/s}\) [1 pt]
Problem 6: Subway Departure & Gain Rate [ 3 Points ]
(a) Final Velocity Calculation: \(v_i = 0\text{ m/s}\), \(a = 1.5\text{ m/s}^2\), \(t = 12.0\text{ s}\)(b) Velocity Gain per Second: The train gained exactly \(1.5\text{ m/s}\) of speed every second (definition of \(\text{m/s}^2\)).
Part a calculation & units [2 pts]
Part b definition of acceleration [1 pt]
Turn to Page 3 for Solutions to Graph Problems 7 through 10 →
Part 2: Graph Solutions Guide Page 3 of 4
Solutions: Graph Problems (7 – 10) Model slope calculations, coordinates, and interpretation keys (12 pts total).
Graph 1 Solutions: Hyperloop Capsule Test Run Questions 7 & 8 [ 6 pts ]
Question 7: Segment A Slope Calculation (\(t = 0\) to \(4\text{ s}\)) [ 3 Points ]
Student reads coordinates: \((t_1, v_1) = (0\text{ s}, 0\text{ m/s})\) and \((t_2, v_2) = (4\text{ s}, 30\text{ m/s})\)
Scoring: Reading coordinates from graph accurately [1 pt]; slope setup [1 pt]; final answer \(+7.5\text{ m/s}^2\) [1 pt].
Question 8: Segment B Acceleration & Motion Analysis (\(t = 4\) to \(10\text{ s}\)) [ 3 Points ]
\(\text{Slope } a_B = \dfrac{30 - 30}{10 - 4} = \dfrac{0}{6} = \mathbf{0\text{ m/s}^2}\)
Reason / Meaning: The graph line is completely flat (horizontal) with zero rise (\(\Delta v = 0\)). This proves the capsule has zero acceleration and cruises at a constant velocity of \(30\text{ m/s}\) .
Scoring: Value \(0\text{ m/s}^2\) [1.5 pts]; explanation referencing horizontal line / constant speed [1.5 pts].
Graph 2 Solutions: Emergency Braking & Reversal Analysis Questions 9 & 10 [ 6 pts ]
Question 9: Slope of Braking Phase (\(t = 0\) to \(4\text{ s}\)) [ 3 Points ]
Student reads coordinates: \((0\text{ s}, 20\text{ m/s})\) and \((4\text{ s}, 0\text{ m/s})\)
Scoring: Reading coordinates from graph accurately [1 pt]; negative numerator preserved [1 pt]; answer \(-5.0\text{ m/s}^2\) [1 pt].
Question 10: Zero Crossing & Direction Reversal Interpretation [ 3 Points ]
(a) Moment of rest: At exactly \(t = 4.0\text{ seconds}\) , the line intercepts the time axis where velocity is \(0\text{ m/s}\).(b) Physical meaning for \(t > 4\text{ s}\): Velocity becomes negative, meaning the vehicle has reversed direction and is now moving backward (speeding up in reverse to \(-10\text{ m/s}\)).
Scoring: Identifying \(t = 4.0\text{ s}\) [1.5 pts]; identifying reversal of motion direction [1.5 pts].
Turn to Page 4 for Solutions to Multi-Stage Drone Graph Problem 11 →
Part 2: Graph Problem 11 Solution Key Page 4 of 4
Solutions: Graph Problem 11 Model responses for multi-stage velocity analysis and comparative slopes (5 pts total).
Graph 3 Solutions: Autonomous Cargo Drone Profile Problem 11 [ 5 Points ]
Part (a): Steepest Descent Slope in Phase 3 (\(t = 10\) to \(12\text{ s}\)) [ 2 Points ]
Coordinates from graph: \((t_1, v_1) = (10\text{ s}, 24\text{ m/s})\) and \((t_2, v_2) = (12\text{ s}, 8\text{ m/s})\)
Scoring: Correct coordinate extraction and math [1 pt]; negative sign and \(\text{m/s}^2\) unit [1 pt].
Part (b): Final Glide Slope in Phase 4 (\(t = 12\) to \(16\text{ s}\)) [ 2 Points ]
Coordinates from graph: \((t_1, v_1) = (12\text{ s}, 8\text{ m/s})\) and \((t_2, v_2) = (16\text{ s}, 0\text{ m/s})\)
Scoring: Correct coordinate identification [1 pt]; final answer \(-2.0\text{ m/s}^2\) with units [1 pt].
Part (c): Comparative Slope Analysis & Line Steepness [ 1 Point ]
Model Explanation: The magnitude of acceleration in Phase 3 (\(|-8.0\text{ m/s}^2|\)) is four times larger than in Phase 4 (\(|-2.0\text{ m/s}^2|\)) because the line in Phase 3 is noticeably steeper . It sheds \(16\text{ m/s}\) in just \(2\text{ seconds}\) (rate of \(-8\text{ m/s}^2\)), whereas Phase 4 spreads an \(8\text{ m/s}\) drop over \(4\text{ seconds}\) (gentle slope of \(-2\text{ m/s}^2\)). Steepness on a \(v\)-\(t\) graph directly indicates the rate of velocity change.
Teacher Quick Reference: All Drone Mission Phases:
Phase 1: \(a = +6.0\text{ m/s}^2\)
Phase 2: \(a = 0\text{ m/s}^2\)
Phase 3: \(a = -8.0\text{ m/s}^2\)
Phase 4: \(a = -2.0\text{ m/s}^2\)
Complete Assignment Key: Total Score = 35 points across 11 multi-tier kinematics problems. ✓ 4-Page Key Complete
