Question Statement

A cyclist rides in a straight line for 20 minutes. He waits for half an hour, then returns in a straight line to his starting point in 15 minutes. The displacement-time graph for his journey is shown below.

(i) Work out the average velocity for each stage of the journey in km/h.

(ii) Write down 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. Average velocity depends on the change in displacement (final position minus initial position), while average speed depends on the total distance traveled regardless of direction.

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Solution

Part (i): Average velocity for each stage

Stage $OA$ (First 20 minutes):

The cyclist moves from displacement $0$ km to $5$ km in 20 minutes.

$V_{\text{av}}=\frac{\text{Change in displacement}}{\text{Time}}$

$=\frac{(5-0)\text{ km}}{20\text{ min}}$

To convert minutes to hours, divide by 60:

$=\frac{5}{\frac{20}{60}}\text{ km/h}$

$=5\times 3\text{ km/h}$

$=15\text{ km/h}$

Stage $AB$ (Next 30 minutes):

The cyclist waits at displacement $5$ km for 30 minutes.

$V_{\text{av}}=\frac{\text{Change in displacement}}{\text{Time}}$

$=\frac{5-5}{30}=0\text{ km/h}$

Since there is no change in position, the velocity is zero.

Stage $BC$ (Final 15 minutes):

The cyclist returns from displacement $5$ km to $0$ km in 15 minutes.

$V_{\text{av}}=\frac{0-5}{15}\frac{\text{km}}{\text{min}}$

Converting minutes to hours:

$=\frac{-5}{\frac{15}{60}}\text{ km/h}$

$=-5\times 4\text{ km/h}$

$=-20\text{ km/h}$

The negative sign indicates the cyclist is moving in the opposite direction (back toward the starting point).

Part (ii): Average velocity for the whole journey

Average velocity depends on the total change in displacement. Since the cyclist returns to the starting point, the final displacement equals the initial displacement ($0$ km).

$V_{\text{av}}\text{ for whole journey}=\frac{\text{Total change in displacement}}{\text{Total time taken}}$

First, calculate total time:

$\text{Total time}=20+30+15=65\text{ min}=\frac{65}{60}\text{ h}$

Calculate average velocity:

$\text{Average velocity}=\frac{0}{\frac{65}{60}}=0\text{ km/h}$

Part (iii): Average speed for the whole journey

Average speed uses total distance traveled (a scalar quantity), not displacement. The cyclist travels $5$ km away and $5$ km back, for a total distance of $10$ km.

$\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$

$=\frac{10}{\frac{65}{60}}$

$=\frac{10\times 60}{65}$

$=\frac{600}{65}=\frac{120}{13}\text{ km/h}$

(Approximately $9.23$ km/h)

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Key Formulas or Methods Used

• Average velocity: $v_{\text{av}}=\frac{\Delta s}{\Delta t}=\frac{\text{Final displacement}-\text{Initial displacement}}{\text{Time taken}}$
• Average speed: ${\text{Speed}}_{\text{av}}=\frac{\text{Total distance traveled}}{\text{Total time taken}}$
• Unit conversion: Minutes to hours: $t_{\text{hours}}=\frac{t_{\text{minutes}}}{60}$
• Interpreting displacement-time graphs:
  • Positive gradient = positive velocity (moving away from origin)
  • Zero gradient = zero velocity (stationary)
  • Negative gradient = negative velocity (moving toward origin)

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Summary of Steps

1. Identify stage parameters from the graph: Determine displacement values and time intervals for segments $OA$, $AB$, and $BC$.
2. Calculate stage velocities using $v=\frac{\Delta s}{\Delta t}$, converting minutes to hours by dividing by 60.
3. Note direction: Assign negative sign to velocity when returning toward the start (stage $BC$).
4. Whole journey velocity: Recognize that total displacement is zero (returns to start), making average velocity zero.
5. Calculate average speed: Use total distance ($10$ km) divided by total time ($\frac{65}{60}$ h), yielding $\frac{120}{13}$ km/h.