Question Statement

This is a displacement-time graph for a car travelling along a straight road. The journey is divided into 5 stages labelled $\mathbf{A}$ to $\mathbf{E}$.

(i) Work out the average velocity for each stage of the journey.

(ii) State the average velocity for the whole journey.

(iii) Work out the average speed for the whole journey.

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Background and Explanation

This problem involves interpreting a displacement-time graph, where the gradient represents velocity. Remember that velocity is a vector quantity (includes direction, hence can be negative), calculated as displacement over time, while speed is a scalar quantity calculated as total distance travelled over time, regardless of direction.

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Solution

Part (i): Average Velocity for Each Stage

Average velocity is calculated using the formula $v=\frac{\Delta s}{\Delta t}$, where $\Delta s$ is the change in displacement and $\Delta t$ is the time interval.

Stage A

During stage A, the car moves from displacement $0$ km to $40$ km.

$\begin{array}{rl}\text{Displacement}& =40-0=40\text{ km}\\ \text{Time}& =10:00-9:30\\ & =30\text{ min}\\ & =\frac{30}{60}\text{ h}\\ & =\frac{1}{2}\text{ h}\\ \text{Velocity}& =\frac{\text{Displacement}}{\text{Time}}\\ & =\frac{40}{1/2}=40\times 2\\ & =80\text{ km/h}\end{array}$

Stage B

During stage B, the car continues moving forward from $40$ km to $60$ km.

$\begin{array}{rl}\text{Displacement}& =60-40\\ & =20\text{ km}\\ \text{Time}& =10:30-10:00\\ & =30\text{ min}\\ & =\frac{30}{60}=\frac{1}{2}\text{ h}\\ \text{Velocity}& =\frac{\text{Displacement}}{\text{Time}}\\ & =\frac{20}{1/2}\\ & =20\times 2=40\text{ km/h}\end{array}$

Stage C

During stage C, the car remains stationary (horizontal line on graph).

$\begin{array}{rl}\text{Displacement}& =60-60\\ & =0\text{ km}\\ \text{Time}& =11:00-10:30\\ & =30\text{ min}\\ & =\frac{30}{60}=\frac{1}{2}\text{ h}\\ \text{Velocity}& =\frac{\text{Displacement}}{\text{Time}}\\ & =\frac{0}{1/2}\\ & =0\text{ km/h}\end{array}$

Stage D

During stage D, the car moves forward from $60$ km to $100$ km.

$\begin{array}{rl}\text{Displacement}& =100-60\\ & =40\text{ km}\\ \text{Time}& =12:00-11:00\\ & =1\text{ h}\\ \text{Velocity}& =\frac{\text{Displacement}}{\text{Time}}\\ & =\frac{40}{1}\\ & =40\text{ km/h}\end{array}$

Note: Based on the graph timeline, Stage D runs from 11:00 to 12:00.

Stage E

During stage E, the car returns to the starting point (displacement decreases from $100$ km to $0$ km).

$\begin{array}{rl}\text{Displacement}& =0-100\\ & =-100\text{ km}\\ \text{Time}& =13:30-12:00\\ & =1\text{ h }30\text{ min}=90\text{ min}=\frac{90}{60}=\frac{3}{2}\text{ h}\\ \text{Velocity}& =\frac{\text{Displacement}}{\text{Time}}\\ & =\frac{-100}{3/2}=\frac{-200}{3}\text{ km/h}\\ & \approx -66.67\text{ km/h}\end{array}$

The negative sign indicates the car is moving in the opposite direction (returning towards the origin).

Part (ii): Average Velocity for the Whole Journey

Average velocity depends on the total displacement (final position minus initial position).

$\text{Average velocity}=\frac{\text{Total displacement}}{\text{Total time}}$

Total displacement $=0-0=0$ km (the car starts and ends at the same position).

$\text{Total time}=13:30-9:30=4\text{ h}$

$\text{Average velocity}=\frac{0}{4}=0\text{ km/h}$

Part (iii): Average Speed for the Whole Journey

Average speed uses the total distance travelled (sum of all distances moved, ignoring direction).

$\begin{array}{rl}\text{Average speed}& =\frac{\text{Total distance}}{\text{Total time}}\\ \text{Total distance}& =40+20+0+40+100\\ & =200\text{ km}\\ \text{Total time}& =13:30-9:30=4\text{ h}\\ \text{Average speed}& =\frac{200}{4}=50\text{ km/h}\end{array}$

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Key Formulas or Methods Used

• Average Velocity: $v=\frac{\Delta s}{\Delta t}=\frac{\text{change in displacement}}{\text{change in time}}$ (vector quantity, can be negative)
• Average Speed: $\text{speed}=\frac{\text{total distance}}{\text{total time}}$ (scalar quantity, always positive)
• Time Conversion: Converting minutes to hours by dividing by 60 (e.g., $30\text{ min}=\frac{1}{2}\text{ h}$)
• Displacement vs Distance: Displacement considers direction (start to end position), while distance sums all movement regardless of direction

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Summary of Steps

1. Identify intervals: Determine the time duration for each stage by subtracting start time from end time, converting minutes to hours.
2. Calculate stage displacements: Find the change in vertical axis values (final displacement minus initial displacement) for each stage.
3. Compute stage velocities: Divide each stage's displacement by its time duration. Note negative values indicate reverse direction.
4. Calculate overall average velocity: Find total displacement (zero for a round trip) and divide by total journey time (4 hours).
5. Calculate overall average speed: Sum all individual distances travelled (200 km) and divide by total time (4 hours).