A: Velocity-Time Graph

Let's break the problem down and create a sketch of the velocity-time graph for the cyclist's motion.

1. Initial Acceleration (0 to 5 seconds): The cyclist starts from rest and accelerates to a constant speed over 5 seconds. This part of the graph is a straight line sloping upwards from the origin, indicating increasing velocity.

2. Constant Speed (5 to 15 seconds): The cyclist maintains constant speed for 10 seconds. This part of the graph is a horizontal straight line, representing constant velocity.

3. Deceleration to Stop (15 to 18 seconds): The cyclist decelerates uniformly to rest in 3 seconds. This part of the graph is a straight line sloping downwards to zero velocity.

The velocity-time graph would look like this:

• A rising straight line from $t = 0$ to $t = 5$ seconds.
• A horizontal line from $t = 5$ to $t = 15$ seconds.
• A falling straight line from $t = 15$ to $t = 18$ seconds.

B: Motion Representation on the Graph

1. Acceleration (0 to 5 seconds): This is the phase where the cyclist's velocity increases. It is represented by an upward-sloping straight line. The slope of this line indicates the rate of acceleration.

2. Constant Speed (5 to 15 seconds): During this phase, the cyclist moves at a constant velocity. It is represented by a horizontal line, showing that the velocity is constant.

3. Deceleration (15 to 18 seconds): The cyclist slows down uniformly to a stop. It is represented by a downward-sloping straight line. The slope of this line indicates the rate of deceleration.

C: Calculation of Total Displacement

The total displacement is given by the area under the velocity-time graph. We'll calculate the area for each phase.

1. Acceleration Phase (0 to 5 seconds):

   • The shape of this part of the graph is a triangle with base 5 seconds and height $v$ (the constant speed reached after acceleration).
   • Area of the triangle: $A_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times v = 2.5v$

2. Constant Speed Phase (5 to 15 seconds):

   • The shape of this part is a rectangle with base 10 seconds and height $v$.
   • Area of the rectangle: $A_2 = \text{base} \times \text{height} = 10 \times v = 10v$

3. Deceleration Phase (15 to 18 seconds):

   • This part of the graph forms a triangle with base 3 seconds and height $v$.
   • Area of the triangle: $A_3 = \frac{1}{2} \times 3 \times v = 1.5v$

Now, the total displacement is the sum of all these areas: $\text{Total displacement} = A_1 + A_2 + A_3 = 2.5v + 10v + 1.5v = 14v$

Hence, the total displacement is $14v$, where $v$ is the cyclist’s maximum velocity during the constant speed phase.

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Would you like to clarify anything or go into more details? Here are some related questions for further exploration:

1. What is the formula for calculating displacement from a velocity-time graph?
2. How does the slope of the graph relate to acceleration?
3. What are the different ways of calculating average velocity for this trip?
4. How would the graph change if the cyclist decelerates non-uniformly?
5. Can this method of calculating displacement be applied if the velocity changes erratically?

Tip: The area under a velocity-time graph always gives the displacement. For uniform acceleration and deceleration, triangles and rectangles help break down the areas.