Comprehensive Study Guide to Distance-Time Modeling: Exercise 18D*

Overview of Distance–Time Graph Construction for Exercise 18D*

• The primary objective of these exercises is to represent the relationship between time and distance visually on a Cartesian plane where time is the independent variable and distance is the dependent variable.
• To construct the graphs as specified in Problem 1, a coordinate system must be established with the following parameters:     - Horizontal Axis (x-axis): Represents Time ($t$) in units of hours ($\text{h}$). The scale must span from a minimum value of $0$ to a maximum value of $3$.     - Vertical Axis (y-axis): Represents Distance ($d$) in units of kilometers ($\text{km}$). The scale must span from a minimum value of $0$ to a maximum value of $110$.

Technical Plotting Guidelines for Constant Speeds

• In a distance–time graph, objects moving at a constant speed are represented by straight lines originating from the origin ($0, 0$) unless otherwise specified. The gradient (slope) of the line represents the speed of the object: $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$.
• The formula used to determine coordinates for plotting is $\text{Distance} = \text{Speed} \times \text{Time}$.
• Scenario (a): Van Travelling at $80\text{ km/h}$     - At $t = 0\text{ h}$, distance $d = 0\text{ km}$.     - At $t = 1\text{ h}$, distance $d = 80\text{ km/h} \times 1\text{ h} = 80\text{ km}$.     - Key Plotting Coordinates: $(0, 0)$ and $(1, 80)$.
• Scenario (b): Cyclist Travelling at $50\text{ km/h}$     - At $t = 0\text{ h}$, distance $d = 0\text{ km}$.     - At $t = 2\text{ h}$, distance $d = 50\text{ km/h} \times 2\text{ h} = 100\text{ km}$.     - Key Plotting Coordinates: $(0, 0)$ and $(2, 100)$.
• Scenario (c): Car Travelling at $100\text{ km/h}$     - At $t = 0\text{ h}$, distance $d = 0\text{ km}$.     - At $t = 1\text{ h}$, distance $d = 100\text{ km/h} \times 1\text{ h} = 100\text{ km}$.     - Key Plotting Coordinates: $(0, 0)$ and $(1, 100)$.
• Scenario (d): Train Travelling at $70\text{ km/h}$     - At $t = 0\text{ h}$, distance $d = 0\text{ km}$.     - At $t = 1\text{ h}$, distance $d = 70\text{ km/h} \times 1\text{ h} = 70\text{ km}$.     - Key Plotting Coordinates: $(0, 0)$ and $(1, 70)$.

Multi-Phase Motion and Stoppage Analysis

• Exercise 2(a) requires the representation of a journey with changing speeds and intervals of rest. This results in a graph composed of multiple line segments with varying gradients.
• Phase 1: Initial Cycling     - Condition: Cycles for $1\text{ hour}$ at a speed of $20\text{ km/h}$.     - Calculation: Distance covered = $20\text{ km/h} \times 1\text{ h} = 20\text{ km}$.     - Plot Segment: A line from $(0, 0)$ to $(1, 20)$.
• Phase 2: Stationary Period (Stopping)     - Condition: Stops for a duration of $1\text{ hour}$.     - Mechanism: During a stop, time continues to advance, but distance remains constant. This is represented by a horizontal line (gradient = $0$).     - Time Calculation: The current timestamp is $1\text{ hour}$; after stopping for $1\text{ hour}$, the timestamp is $2\text{ hours}$.     - Plot Segment: A horizontal line from $(1, 20)$ to $(2, 20)$.
• Phase 3: Final Cycling     - Condition: Continues to cycle for another $2\text{ hours}$ at a speed of $15\text{ km/h}$.     - Calculation of Interval Distance: Distance added = $15\text{ km/h} \times 2\text{ h} = 30\text{ km}$.     - Calculation of Cumulative Totals:         - Total Time: Current time ($2\text{ h}$) + additional duration ($2\text{ h}$) = $4\text{ hours}$.         - Total Distance: Previous distance ($20\text{ km}$) + additional distance ($30\text{ km}$) = $50\text{ km}$.     - Plot Segment: A line from $(2, 20)$ to $(4, 50)$.