Question Statement

A helicopter leaves Islamabad at 09:00. It flies for 45 minutes at $80\mathrm{km}/\mathrm{h}$. It lands for 20 minutes. The helicopter then returns to its base in Islamabad, flying at $100\mathrm{km}/\mathrm{h}$. Draw a displacement-time graph to show the journey.

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Background and Explanation

A displacement-time graph plots the displacement (distance from the starting point) on the vertical axis against time on the horizontal axis. The gradient of the line represents velocity—positive for moving away from the origin, zero when stationary, and negative when returning toward the origin.

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Solution

To construct the displacement-time graph, we calculate the helicopter's position at each key time interval to determine the coordinates of the graph.

Stage 1: Outbound Journey (09:00 to 09:45)

The helicopter travels for 45 minutes at $80\mathrm{km}/\mathrm{h}$ away from Islamabad.

First, convert the time to hours: $t_1=45\text{ minutes}=\frac{45}{60}\text{ hours}=0.75\text{ hours}$

Calculate the displacement from Islamabad: $s=v\times t=80\times 0.75=60\text{ km}$

Determine the arrival time: $09:00+45\text{ minutes}=09:45$

Graph segment: A straight line from coordinate $(09:00,0)$ to $(09:45,60)$ with a positive gradient representing $80\mathrm{km}/\mathrm{h}$.

Stage 2: Stationary Period (09:45 to 10:05)

The helicopter lands and remains stationary for 20 minutes at a constant displacement of $60\text{ km}$ from Islamabad.

Calculate the end time: $09:45+20\text{ minutes}=10:05$

Graph segment: A horizontal line from $(09:45,60)$ to $(10:05,60)$ with zero gradient (velocity = 0).

Stage 3: Return Journey (10:05 to 10:41)

The helicopter returns to Islamabad (displacement returns to $0$) at $100\mathrm{km}/\mathrm{h}$.

Calculate the time required to cover the $60\text{ km}$ return distance: $t_2=\frac{\text{distance}}{\text{speed}}=\frac{60}{100}=0.6\text{ hours}$

Convert the return time to minutes: $t_2=0.6\times 60=36\text{ minutes}$

Determine the final arrival time: $10:05+36\text{ minutes}=10:41$

Graph segment: A straight line from $(10:05,60)$ to $(10:41,0)$ with a negative gradient of $-100$ representing the return velocity. Note that this segment is steeper than the outbound segment because $100>80$.

Final Graph

The complete displacement-time graph consists of three line segments: an upward slope (outbound), a horizontal plateau (landing), and a downward slope (return) that is steeper than the first. The vertical axis shows displacement from Islamabad in kilometers, and the horizontal axis shows time from 09:00 to 10:41.

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Key Formulas or Methods Used

• $\text{Distance}=\text{speed}\times \text{time}$ ($s=vt$)
• $\text{Time}=\frac{\text{distance}}{\text{speed}}$ ($t=\frac{s}{v}$)
• Time conversion: $1\text{ hour}=60\text{ minutes}$
• Gradient of displacement-time graph equals velocity

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Summary of Steps

1. Calculate outbound displacement: $80\times 0.75=60$ km, reaching this position at 09:45
2. Account for stationary period: Maintain 60 km displacement from 09:45 to 10:05
3. Calculate return journey time: $60÷100=0.6$ hours = 36 minutes, arriving at 10:41
4. Set up axes: Time (horizontal) scaled from 09:00 to approximately 10:45, Displacement (vertical) from 0 to 60 km
5. Plot key points: $(09:00,0)\to (09:45,60)\to (10:05,60)\to (10:41,0)$
6. Connect with straight lines: First segment gradient = 80, second = 0, third = -100 (steeper descent)