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Let's tackle this problem step by step.

### (a) Find the distance \( AB \).

To find the distance \( AB \), we need to calculate the area under the speed-time graph, as the area under such a graph gives the total distance traveled.

The speed-time graph is divided into three sections:
1. **Acceleration phase (0 to 5 seconds)**: A triangular section.
2. **Constant speed phase (5 to 20 seconds)**: A rectangular section.
3. **Deceleration phase (20 to 30 seconds)**: Another triangular section.

#### 1. Acceleration Phase (0 to 5 seconds):
- This is a triangle with a base of 5 seconds and a height of 5 m/s.
- The area of a triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \).
  \[
  \text{Area}_1 = \frac{1}{2} \times 5 \, \text{s} \times 5 \, \text{m/s} = 12.5 \, \text{m}
  \]

#### 2. Constant Speed Phase (5 to 20 seconds):
- This is a rectangle with a base of \( 20 - 5 = 15 \) seconds and a height of 5 m/s.
- The area of a rectangle is \( \text{base} \times \text{height} \).
  \[
  \text{Area}_2 = 15 \, \text{s} \times 5 \, \text{m/s} = 75 \, \text{m}
  \]

#### 3. Deceleration Phase (20 to 30 seconds):
- This is another triangle with a base of \( 30 - 20 = 10 \) seconds and a height of 5 m/s.
  \[
  \text{Area}_3 = \frac{1}{2} \times 10 \, \text{s} \times 5 \, \text{m/s} = 25 \, \text{m}
  \]

#### Total Distance:
The total distance \( AB \) is the sum of these three areas:
\[
\text{Total Distance} = 12.5 \, \text{m} + 75 \, \text{m} + 25 \, \text{m} = 112.5 \, \text{m}
\]

### (b) Sketch a distance-time graph for the journey of the car from A to B.

The distance-time graph for this journey should follow these key features:
1. **First 5 seconds**: The car is accelerating uniformly, so the graph will be a curve that becomes steeper, starting from 0 and reaching a distance of 12.5 meters at 5 seconds.
2. **Next 15 seconds (5 to 20 seconds)**: The car is moving at a constant speed, so the graph will be a straight line with a constant slope. The total distance at 20 seconds is 87.5 meters (12.5 m + 75 m).
3. **Last 10 seconds (20 to 30 seconds)**: The car decelerates uniformly, so the graph will again be a curve, but this time it will flatten as the car comes to rest at 112.5 meters after 30 seconds.

Would you like more details or clarifications on any part?

Here are five related questions to expand understanding:
1. How is the total distance calculated using a velocity-time graph?
2. What would the shape of the distance-time graph look like for a uniformly accelerating object?
3. How would the graph change if the car had different acceleration and deceleration rates?
4. What would be the effect on the distance if the car traveled for a longer time at constant speed?
5. How can the average speed of the journey be calculated?

**Tip:** The area under a velocity-time graph always gives the distance traveled, which is a useful tool in kinematics problems!

Figure 1 shows the speed-time graph for the journey of a car moving in a long queue of traffic on a straight horizontal road. The car is initially at rest and undergoes acceleration, constant speed, and deceleration phases. (a) Find the distance AB. (b) Sketch a distance-time graph for the journey of the car from A to B.

Solve a kinematics problem by finding the total distance traveled by a car based on its speed-time graph. Learn how to calculate distance using areas under the graph and sketch a distance-time graph for uniform acceleration, constant speed, and deceleration phases. Suitable for high school students (Grades 9-11).

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