Question Statement

Ahmed leaves home at 11:00 am. He cycles at a speed of $16\mathrm{km}/\mathrm{h}$ for 90 minutes. He stops for half an hour. Ahmed then cycles home and arrives at $3\mathbf{pm}$.

(i) Draw a displacement-time graph to show Ahmed's journey.

(ii) What is Ahmed's average speed on the return part of his cycle?

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Background and Explanation

This problem involves analyzing motion with constant speed, calculating time intervals, and interpreting displacement-time graphs. You need to track Ahmed's position relative to home over time, noting that displacement represents distance from the starting point and returns to zero when he arrives back home.

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Solution

Part (i): Displacement-Time Graph

To construct the displacement-time graph, we first calculate the key time points and distances:

Outbound journey:

• Start time: 11:00 am ($t=0$)
• Speed: $16\mathrm{km}/\mathrm{h}$
• Duration: $90\text{ minutes}=1.5\text{ hours}$
• Distance from home: $16\times 1.5=24\mathrm{km}$

Rest period:

• Start time: 11:00 am + 1.5 hours = 12:30 pm
• Duration: 30 minutes = 0.5 hours
• End time: 12:30 pm + 0.5 hours = 1:00 pm (13:00)
• During rest, displacement remains constant at $24\mathrm{km}$

Return journey:

• Start time: 1:00 pm (13:00)
• End time: 3:00 pm (15:00)
• Duration: 2 hours
• Final displacement: $0\mathrm{km}$ (back home)

The graph consists of three segments: a straight line with positive slope from (11:00, 0) to (12:30, 24), a horizontal line from (12:30, 24) to (13:00, 24), and a straight line with negative slope from (13:00, 24) to (15:00, 0).

Part (ii): Average Speed on Return Journey

First, determine the distance Ahmed must travel to return home. From part (i), he traveled $24\mathrm{km}$ away from home, so he must cover the same distance to return:

$\text{Distance}=24\mathrm{km}$

Next, calculate the time taken for the return journey. Ahmed departs on his return at 1:00 pm (after cycling 90 minutes and resting 30 minutes from 11:00 am) and arrives home at 3:00 pm:

$\text{Time}=3:00-1:00=2\text{ hours}$

Using the formula for average speed:

$v_{av}=\frac{\text{Distance}}{\text{Time}}=\frac{24}{2}=12\mathrm{km}/\mathrm{h}$

Therefore, Ahmed's average speed on the return part of his cycle is $12\mathrm{km}/\mathrm{h}$.

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Key Formulas or Methods Used

• $\text{Distance}=\text{Speed}\times \text{Time}$
• $\text{Average Speed}=\frac{\text{Total Distance}}{\text{Total Time}}$
• Displacement-time graph interpretation: slope represents velocity, horizontal line represents rest (zero velocity)
• Time unit conversions: $90\text{ min}=1.5\text{ h}$ and $30\text{ min}=0.5\text{ h}$

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Summary of Steps

1. Calculate outbound distance: $16km/h\times 1.5\mathrm{h}=24\mathrm{km}$ from home
2. Determine timeline: Depart 11:00, reach destination 12:30, rest until 13:00, arrive home 15:00
3. Draw displacement-time graph: Rising line (11:00–12:30), horizontal line (12:30–13:00), falling line (13:00–15:00)
4. Identify return distance: $24\mathrm{km}$ (same as outbound)
5. Calculate return time: $15:00-13:00=2$ hours
6. Compute average speed: $\frac{24\mathrm{km}}{2\mathrm{h}}=12km/h$