Question Statement

Fatima left home at 10:00 and cycled north-east in a straight line. The diagram shows a displacement-time graph for her journey.

(i) Find Fatima's velocity between 10:00 and 11:00.

(ii) Estimate the time that Fatima passed her home.

(iii) Find Fatima's velocity for each of the last two stages of her journey.

(iv) Calculate Fatima's average speed for her entire journey.

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Background and Explanation

This problem requires interpreting a displacement-time graph, where the gradient (slope) of the line represents velocity. Remember that displacement is a vector quantity (can be positive or negative relative to the starting point), while distance is always positive. Velocity includes directional information indicated by the sign of the slope, whereas speed uses the absolute value of distance traveled.

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Solution

Part (i): Velocity between 10:00 and 11:00

During the first hour, Fatima travels from home (displacement = 0) to a displacement of 12 km. Velocity is calculated as the change in displacement divided by the time taken.

$\begin{array}{rl}\text{Displacement}& =12-0=12\text{ km}\\ \text{Time}& =11:00-10:00=1\text{ h}\\ \text{Velocity}& =\frac{\text{Displacement}}{\text{Time}}\\ & =\frac{12}{1}=12\text{ km/h}\end{array}$

Fatima's velocity between 10:00 and 11:00 is $12\text{ km/h}$ in the north-east direction.

Part (ii): Time Fatima passed her home

Fatima passes her home when her displacement returns to zero. On the graph, this occurs where the curve intersects the time axis (horizontal axis) during her return journey. Observing the graph, this intersection point appears approximately halfway between 12:00 and 13:00.

Time that Fatima passed her home is nearly 12:30.

Part (iii): Velocity for the last two stages

The return journey consists of two distinct stages after Fatima initially turned back. Note that between 11:00 and 11:30, she remained stationary at the 12 km mark (indicated by a horizontal line on the graph), before beginning her return.

First return stage (11:30 to 13:00): Fatima travels from the 12 km position back past her home to a position 4 km on the opposite side (displacement = -4 km).

$\begin{array}{rl}\text{Displacement}& =-4-12=-16\text{ km}\\ \text{Time}& =13:00-11:30=1.5\text{ h}\\ \text{Velocity}& =\frac{\text{Displacement}}{\text{Time}}\\ & =\frac{-16}{1.5}=\frac{-32}{3}\text{ km/h}\end{array}$

Second return stage (13:00 to 14:00): Fatima travels from the -4 km position back toward her home.

$\begin{array}{rl}\text{Displacement}& =4-(-4)=4\text{ km}\\ \text{Time}& =14:00-13:00=1\text{ h}\\ \text{Velocity}& =\frac{\text{Displacement}}{\text{Time}}\\ & =\frac{4}{1}=4\text{ km/h}\end{array}$

Thus, the velocities for the last two stages are $-\frac{32}{3}\text{ km/h}$ (or approximately $-10.67\text{ km/h}$) and $4\text{ km/h}$ respectively. The negative sign indicates motion in the opposite direction to the initial north-east journey.

Part (iv): Average speed for the entire journey

Average speed uses total distance traveled (scalar sum of all path lengths) divided by total time. The journey consists of: 12 km outbound, a stationary period (0 km), 16 km returning past home, and 4 km returning to home.

$\begin{array}{rl}\text{Total distance}& =12+0+16+4\\ & =32\text{ km}\\ \text{Total time}& =14:00-10:00=4\text{ h}\\ \text{Average speed}& =\frac{\text{Total distance}}{\text{Total time}}\\ & =\frac{32}{4}=8\text{ km/h}\end{array}$

Fatima's average speed for her entire journey is $8\text{ km/h}$.

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Key Formulas or Methods Used

• Velocity from displacement-time graph: $v=\frac{\Delta s}{\Delta t}=\frac{\text{final displacement}-\text{initial displacement}}{\text{time elapsed}}$
• Average speed calculation: $\text{Average speed}=\frac{\text{Total distance traveled}}{\text{Total time taken}}$
• Interpreting graph features: Zero displacement indicates the object is at the starting point (home); horizontal lines indicate stationary periods (zero velocity)
• Vector nature of displacement: Negative displacement indicates position in the opposite direction from the initial travel direction

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Summary of Steps

1. Calculate initial velocity: Determine the gradient of the first segment: $\frac{12\text{ km}-0}{1\text{ h}}=12\text{ km/h}$
2. Identify home passage time: Locate where the displacement returns to zero on the graph (approximately 12:30)
3. Calculate return stage velocities:
   • Stage 1 (11:30–13:00): $\frac{-4-12}{1.5}=-\frac{32}{3}\text{ km/h}$
   • Stage 2 (13:00–14:00): $\frac{4-(-4)}{1}=4\text{ km/h}$
4. Compute average speed: Sum all distances including the stationary period ($12+0+16+4=32$ km), divide by total duration (4 hours) to obtain $8\text{ km/h}$