Question Statement

Dabeer leaves at 14:00. He drives at an average speed of $14\mathrm{km}/\mathrm{h}$ for $3\frac{1}{2}$ hours. Dabeer stops for 30 minutes. He then drives home at $70\mathrm{km}/\mathrm{h}$. Draw a displacement-time graph to show Dabeer's journey.

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Background and Explanation

A displacement-time graph tracks an object's position relative to its starting point over time. The gradient (slope) of the line represents velocity—positive when moving away, zero when stationary, and negative when returning. To construct the graph, you must calculate the displacement at each stage using the relationship between speed, distance, and time.

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Solution

To draw the displacement-time graph, we calculate the displacement and time at each key point of the journey.

Stage 1: Outbound Journey (14:00 to 17:30)

Dabeer drives away from home at a constant speed of $14\mathrm{km}/\mathrm{h}$ for $3.5$ hours.

Calculate the displacement from home: $\text{Displacement}=\text{Speed}\times \text{Time}=14\times 3.5=49\text{ km}$

Calculate the arrival time at the destination: $14:00+3.5\text{ hours}=14:00+3\text{ hours }30\text{ minutes}=17:30$

Graph segment: A straight line from coordinates $(14:00,0)$ to $(17:30,49)$ with a positive gradient of $14$.

Stage 2: Stationary Period (17:30 to 18:00)

Dabeer stops for $30$ minutes. During this time, his displacement remains unchanged at $49\mathrm{km}$ from home.

Calculate the time when he starts driving again: $17:30+30\text{ minutes}=18:00$

Graph segment: A horizontal line from $(17:30,49)$ to $(18:00,49)$, indicating zero velocity.

Stage 3: Return Journey (18:00 to 18:42)

Dabeer drives home at $70\mathrm{km}/\mathrm{h}$, covering the same distance of $49\mathrm{km}$ back to the starting point.

Calculate the time required for the return trip: $\text{Time}=\frac{\text{Distance}}{\text{Speed}}=\frac{49}{70}=0.7\text{ hours}$

Convert decimal hours to minutes: $0.7\times 60=42\text{ minutes}$

Calculate the final arrival time back home: $18:00+42\text{ minutes}=18:42$

Graph segment: A straight line from $(18:00,49)$ to $(18:42,0)$ with a negative gradient of $-70$ (negative because displacement is decreasing).

Completed Graph

The displacement-time graph should have:

• Horizontal axis: Time (scaled from 14:00 to at least 19:00)
• Vertical axis: Displacement in km (scaled from 0 to 50 km)
• Three connected line segments showing: outward travel (rising), stop (flat), and return travel (falling to zero)

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Key Formulas or Methods Used

• $\text{Distance}=\text{Speed}\times \text{Time}$ or $d=s\times t$
• $\text{Time}=\frac{\text{Distance}}{\text{Speed}}$ or $t=\frac{d}{s}$
• Gradient of a displacement-time graph represents velocity
• Horizontal line indicates zero velocity (stationary object)

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Summary of Steps

1. Calculate outbound displacement: $14\times 3.5=49$ km
2. Determine arrival time at destination: $14:00+3.5$ hours $=17:30$
3. Add stationary period: $17:30+30$ minutes $=18:00$
4. Calculate return journey duration: $\frac{49}{70}=0.7$ hours $=42$ minutes
5. Determine final arrival time: $18:00+42$ minutes $=18:42$
6. Plot the four key points: $(14:00,0)$, $(17:30,49)$, $(18:00,49)$, and $(18:42,0)$
7. Connect with straight lines: positive gradient ($14$), zero gradient (horizontal), negative gradient ($-70$)