Distance-Time Graphs Comprehensive Study Notes

Interpretation of the Gradient: The gradient (slope) of a distance-time graph represents the speed of an object. A steeper line indicates a higher speed, while a shallower line represents a slower speed.
Horizontal Lines: A flat horizontal section on a distance-time graph indicates that the distance from the starting point is not changing; therefore, the object is stationary or at rest (e.g., during a meeting or a stop at a shop).
Returning to Start: A line sloping downwards towards the x-axis indicates that the object is moving back towards its starting position.
Mathematical Formula for Speed: $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$

Journey Overview: Clive drove from home to an office for a meeting and then drove straight back home.
Key Graph Points:
- Departure: Clive leaves home at $09:00$ .
- Arrival at Office: Clive reaches the office at $11:00$ .
- Distance to Office: The office is $5\,km$ away from Clive's home.
- Duration of Journey to Meeting: It took Clive $2\text{ hours}$ to drive to the meeting (from $09:00$ to $11:00$ ).
- The Meeting: The meeting lasted for $2\text{ hours}$ , indicated by the horizontal line from $11:00$ to $13:00$ .
- Return Journey: Clive leaves the office at $13:00$ and arrives back home at approximately $14:00$ .

Journey Overview: Danny went for a run, with the distance from home measured in metres over a duration of $60\text{ minutes}$ .
Graph Scales:
- Y-axis: Distance from home (metres), ranging from $0$ to $5000\,m$ .
- X-axis: Time (minutes), ranging from $0$ to $60\text{ minutes}$ .
Data Extraction: After $16\text{ minutes}$ , the distance Danny had run must be interpreted from the graph. Based on a linear progression from $0$ to $20\text{ minutes}$ leading to a certain distance (e.g., if the graph hits $2000\,m$ at $20\text{ minutes}$ , increments occur every $4\text{ minutes}$ per block).

Journey Components: Erica cycled to a friend's house, stopped at shops for snacks, watched a movie, and then cycled back.
Chronological Events:
- Departure: Erica leaves home at $11:00$ .
- Stop at Shops: Erica stops at the shops on the way. The duration of this stop is determined by the length of the first horizontal segment on the graph.
- Arrival at Friend's House: Erica arrives at her friend's house at the time indicated by the start of the second horizontal segment (the movie).
- Total Cycling Time: This is calculated by summing the durations of the non-horizontal segments of the graph.

Journey Parameters:
- Total Distance: The maximum distance reached on the y-axis is $5000\,metres$ , which is equivalent to $5\,km$ .
- Total Time: The journey duration on the x-axis is $140\text{ minutes}$ .
Average Speed Calculation:
- First, convert the total time into hours: $\text{Time in hours} = \frac{140}{60} = \frac{7}{3}\text{ hours}$
- Then, apply the speed formula: $\text{Average Speed} = \frac{5\,km}{\frac{7}{3}\,h} = 5 \times \frac{3}{7} = \frac{15}{7} \approx 2.14\,km/h$

Journey Overview: Grace took a train from Bristol to London for a meeting and returned directly afterward.
Distance and Time Details:
- Travel to London: Grace travels from $07:00$ to approximately $09:00$ . The distance from Bristol to London is $120\,miles$ .
- The Meeting: The meeting is represented by the horizontal line. It starts at $09:00$ and ends at $13:00$ , totaling a duration of $4\text{ hours}$ .
- Leaving the Meeting: Grace departs the meeting at $13:00$ .
- Position at 15:15: To find her distance from home at $15:15$ , identify the y-value corresponding to the time on the return leg of the graph.
- Intermediate Travel: To find the miles traveled between $07:00$ and $08:30$ , calculate the difference in distance values at those two specific timestamps.
Average Speed Calculations:
- Bristol to London Speed: $\text{Speed}_{out} = \frac{120\,miles}{2\text{ hours}} = 60\,mph$
- London to Bristol Speed: Calculate based on the travel time from $13:00$ until her arrival back at distance $0$ .

18:10 – 18:30: Hayley runs $2500\,metres$ . This is a straight line from $(18:10, 0)$ to $(18:30, 2500)$ .
18:30 – 18:40: Rest period. A horizontal line from $(18:30, 2500)$ to $(18:40, 2500)$ .
18:40 – 19:00: Runs a further $1500\,metres$ ( $2500 + 1500 = 4000\,metres$ from home). Line from $(18:40, 2500)$ to $(19:00, 4000)$ .
19:00 – 19:20: Rest period of $20\text{ minutes}$ . Horizontal line from $(19:00, 4000)$ to $(19:20, 4000)$ .
19:20 – 20:10: Runs back home at a steady speed. Line from $(19:20, 4000)$ to $(20:10, 0)$ .

Commute Details:
- Morning Leg: Departs $07:40$ , arrives $08:10$ . Distance = $8\,miles$ .
- Work Duration: Stationary at $8\,miles$ from $08:10$ until $15:30$ .
- Return Trip (Part 1): Departs work at $15:30$ , arrives at a café $5\,miles$ from home at $15:50$ .
- Café Stop: Stays for $1\text{ hour and } 20\text{ minutes}$ ( $15:50$ to $17:10$ ).
- Return Trip (Part 2): Departs café at $17:10$ , arrives home at $17:50$ .
Average Speed Calculation (Morning):
- Time = $30\text{ minutes} = 0.5\text{ hours}$ .
- Speed = $\frac{8\,miles}{0.5\text{ hours}} = 16\,mph$ .

Outward Journey: Graph shows the drive to school and the wait for her daughter.
Return Journey: Jayne drives back home at a steady speed of $27\,km/h$ .
Completing the Graph: If the school is distance $D$ from home, the time taken for the return leg is calculated as: $\text{Time} = \frac{\text{Distance}}{27\,km/h}$
The return line is drawn from the end of the waiting period back to distance $0$ over that calculated time duration.

First Leg: Runs $3000\,metres$ in $8\text{ minutes}$ .
Second Leg: Remaining distance = $5000 - 3000 = 2000\,metres$ .
Speed in Final Leg: $250\,metres\,per\,minute$ .
Time for Final Leg: $\text{Time} = \frac{2000\,m}{250\,m/min} = 8\text{ minutes}$
Graph Plotting:
- Line 1: $(0, 0)$ to $(8, 3000)$ .
- Line 2: $(8, 3000)$ to $(16, 5000)$ .

Common Route: Total distance = $6\,km$ .
Lucas’s Journey:
- Departure time: $08:15$ .
- Speed: $12\,km/h$ .
- Travel Time: $\frac{6\,km}{12\,km/h} = 0.5\text{ hours} = 30\text{ minutes}$ .
- Arrival time: $08:15 + 30\text{ minutes} = 08:45$ .
- Plot Lucas: Straight line from $(08:15, 0)$ to $(08:45, 6)$ .
Analysis: The time Lucas cycles past Kristina is indicated by the intersection of their respective lines on the graph.
Kristina’s Average Speed: Calculated by total distance ( $6\,km$ ) divided by her total time taken as shown on the provided graph.

Total Distance: $18\,miles$ .
Total Time: $60\text{ minutes} = 1\text{ hour}$ .
Average Speed Calculation: $\text{Average Speed} = \frac{18\,miles}{1\text{ hour}} = 18\,mph$