| Constant Flat Horizontal Line |
| Velocity has a single steady value. | Flat Line at Exactly Zero |
| No acceleration (velocity is not changing). |
| Flat Horizontal Line (→) | Flat Line at Exactly Zero |
| Object is stationary (at rest). | Flat Line at Exactly Zero |
| No acceleration (velocity is not changing). |
Unit 1: Kinematics — Graph Conversions Assignment Page 1 of 8
Assignment: Part 1A
Name:
Date:
Task: Curves vs. Lines. Below is a position-time graph of a rocket car. Interval A is a curve (acceleration) starting from rest (v₀ = 0). Interval B is a straight diagonal line (constant speed). Calculate the velocity values and complete the table.
GIVEN GRAPH: Position vs. Time (x vs. t) x (m) t (s) +24 +8 0 4 8
| Time Interval | Detailed Step-by-Step Calculation Workspace
(Show formulas, coordinates, and math steps clearly) | Final Velocity (v) |
| --- | --- | --- |
| Interval A
0 s to 4 s
(Curved Line) |
1. Curve means changing velocity (acceleration). Solve for constant acceleration \(a\) using \(x = \frac{1}{2} a t^2\):
2. Solve for final velocity at t = 4 s using \(v_f = a \cdot t\):
| Rises from
0 to _____ m/s |
| Interval B
4 s to 8 s
(Straight Line) |
1. Straight line means constant velocity. Solve for rise-over-run directly:
| Constant at
_____ m/s |
Unit 1: Kinematics — Graph Conversions Assignment Page 2 of 8
Assignment: Part 1B
Name:
Date:
Using your values from Page 2, draw the resulting velocity segments. (Interval A is linear diagonal rise; Interval B is flat horizontal).
v (m/s) t (s) +4 +2 0 4 8
| Interval | Points on Velocity Graph | Algebraic Slope Calculation: a = (v₂ - v₁)/(t₂ - t₁) | Acceleration (a) |
|---|---|---|---|
| Interval A | |||
| 0 s to 4 s | |||
| (Linear Diagonal) | Start: (0, 0) | ||
| End: (4, ____) | a = (____ - 0) / (4 - 0) = | ||
| a = ________________ m/s² | _____ m/s² | ||
| Interval B | |||
| 4 s to 8 s | |||
| (Flat Horizontal) | Start: (4, ____) | ||
| End: (8, ____) | a = (____ - ____) / (____ - ____) = | ||
| a = ________________ m/s² | _____ m/s² |
Unit 1: Kinematics — Graph Conversions Assignment Page 3 of 8
Assignment: Part 1C
Name:
Date:
Using your values from Page 3, plot constant horizontal segments. (Draw dashed vertical boundaries to show transitions).
a (m/s²) t (s) +2.0 +1.0 0.0 -1.0 -2.0 4 8
1. Explain physically why the curved segment on the Position graph (Interval A) resulted in a diagonal sloped line on the Velocity graph.
2. Interval B features constant velocity. Explain why the resulting acceleration value for this interval is exactly zero, despite the object moving rapidly at 4 m/s.
Unit 1: Kinematics — Graph Conversions Assignment Page 4 of 8
Part 2: Concept Guide
THE AREA PATHWAY: TRANSLATING ACCELERATION TO VELOCITY TO POSITION (a → v → x)
Area Pathway
Shifting Up the Ladder (Using Area Under Curve)
Acceleration a (m/s²)
← area ←
Velocity v (m/s)
← area ←
Position x (m)
Moving up the ladder is an accumulation process. Definite integrals (the geometric area under a curve bound by the time axis) calculate the total change (Δv or Δx) during an interval.
Crucial Concept: Area only tells us the change in value. To plot coordinates, we must know the initial value (initial velocity v₀ or initial position x₀) and add the calculated change:
1. Velocity accumulation equation: v_final = v_initial + Area under Acceleration Graph
2. Position accumulation equation: x_final = x_initial + Area under Velocity Graph
Rectangle Area
Area = Base × Height
Used during periods of constant acceleration or constant velocity (flat lines).
Triangle Area
Area = ½ × Base × Height
Used when velocity changes linearly (diagonal lines representing constant acceleration).
Unit 1: Kinematics — Graph Conversions Assignment Page 5 of 8
Assignment: Part 2A
Name:
Date:
Task: Area Pathway Integration. Below is an acceleration vs. time graph. At t = 0 s, the object has an initial velocity of v₀ = -2 m/s and an initial position of x₀ = +4 m. Calculate the change in velocity (Δv) and final velocity values for each interval.
GIVEN GRAPH: Acceleration vs. Time (a vs. t) a (m/s²) t (s) +1.5 0.0 4 8
| Interval | Area Under Acceleration Graph Workspace
(Calculate Change: Δv = base × height) | Final Velocity (v_f) |
| --- | --- | --- |
| Interval A
0 s to 4 s |
1. Calculate change in velocity (Δv₁ = Area):
2. Add change to initial velocity (v₀ = -2 m/s) to find final velocity:
| v(4) =
________ m/s |
| Interval B
4 s to 8 s |
1. Calculate change in velocity (Δv₂ = Area):
2. Add change to v(4) to find final velocity v(8):
| v(8) =
________ m/s |
Unit 1: Kinematics — Graph Conversions Assignment Page 6 of 8
Assignment: Part 2B
Name:
Date:
Using your values from Page 6, plot the resulting velocity curves. (Starts at -2 m/s, diagonal rise to v_final, then flat horizontal).
v (m/s) t (s) +4 0 -2 4 8
| Interval | Area Calculation & Accumulation Workspace
(Rectangle Area: b × h • Trapezoid/Triangle: [(v_i + v_f)/2] × t) | Final Position (x_f) |
| --- | --- | --- |
| Interval A
0 s to 4 s |
1. Calculate change in position (Δx₁ = Trapezoid Area):
2. Accumulate Position: \(x(4) = x_0 + \Delta x_1\):
| x(4) =
________ m |
| Interval B
4 s to 8 s |
1. Calculate change in position (Δx₂ = Rectangle Area):
2. Accumulate Position: \(x(8) = x(4) + \Delta x_2\):
| x(8) =
________ m |
Unit 1: Kinematics — Graph Conversions Assignment Page 7 of 8
Assignment: Part 2C
Name:
Date:
Plot the points from Page 7. Connect Interval A with correct upward curvature, and Interval B with a straight diagonal line.
x (m) t (s) +24 +16 +8 0 -2 -8 4 8
1. Why was it necessary to be given initial velocity v₀ = -2 m/s and initial position x₀ = +4 m in order to plot the exact coordinates on your graphs? What would happen if we didn't know these?
2. In Interval A, velocity starts negative (-2 m/s) and ends positive (+4 m/s). Explain how the position changes during this interval (does it move backward first, or go straight forward?).
Unit 1: Kinematics — Graph Conversions Assignment Page 8 of 8
• Graphical Connection: This calculates the area of the trapezoid bound under the v-t graph!
Unit 1: Kinematics — Graph Conversions Guide Page 1 of 2
Equations Reference
PART 2 — QUADRATIC MOTION AND THE TIMELESS FORMULA
Analytical Motion
Equation No. 3: x_f = x_i + v_i • t + ½ a • t² • (Quadratic Position)
The Concept: This quadratic equation predicts where an object will land. It separates position into three stacked contributions: where the object starts (x_i), how far it would travel at a constant speed (v_i • t), and the extra quadratic distance added by accelerating (½ a • t²).
• Graphical Connection: This describes the curved parabolic lines drawn on the position-time graph!
Equation No. 4: v_f² = v_i² + 2 a • Δx • (The Timeless Formula)
The Concept: By algebraically combining Equation 1 and Equation 2, we can eliminate the time variable (t) entirely. This timeless relationship is extremely powerful when we only know spatial coordinates and velocities, but have no stop-watch readings.
• Useful Shortcut: Always select this formula when time is completely omitted from the prompt.
The Problem Scenario: A high-speed maglev train is hovering at rest at the station. When given the clearance signal, the train accelerates down a straight track at a constant rate of a = +3.0 m/s² over a distance of Δx = 150 m. Find the final velocity of the train as it passes the 150-meter mark.
Step 1: Identify Given and Unknown Variables
v_i = 0 m/s (from rest) • a = +3.0 m/s² • Δx = 150 m • Unknown: v_f = ?
Step 2: Choose the Correct Formula
We need to solve for v_f. We are given v_i, a, and Δx. Time (t) is completely missing. Looking at our cheat sheet, Equation No. 4 is the perfect fit.
Step 3: Substitute Values and Solve Algebraically
v_f² = v_i² + 2 a • Δx
v_f² = (0)² + 2 • (3.0 m/s²) • (150 m)
v_f² = 0 + 900 → v_f = √900
v_f = +30 m/s
Unit 1: Kinematics — Graph Conversions Guide Page 2 of 2
