https://file.aiquickdraw.com/mathbot/file/1726512510639jtsstn5o.jpg

Let's analyze the provided graph and solve the questions.

### Graph Interpretation
- The graph represents velocity (v) in km/h as a function of time (t) in minutes.
- Key points:
  - From \( t = 0 \) to \( t = 0.5 \) min, the velocity increases from 0 to 45 km/h.
  - From \( t = 0.5 \) to \( t = 2 \) min, the velocity remains constant at 45 km/h.
  - From \( t = 2 \) to \( t = 2.5 \) min, the velocity increases to 75 km/h.
  - From \( t = 2.5 \) to \( t = 4 \) min, the velocity remains constant at 75 km/h.
  - From \( t = 4 \) to \( t = 4.5 \) min, the velocity decreases to 0 km/h.

### (a) Sketch of Acceleration and Distance Graphs
1. **Acceleration Graph:**
   - Acceleration is the change in velocity over time.
   - From \( t = 0 \) to \( t = 0.5 \) min: The car accelerates from 0 to 45 km/h. The slope indicates positive acceleration.
   - From \( t = 0.5 \) to \( t = 2 \) min: No change in velocity, so acceleration is 0.
   - From \( t = 2 \) to \( t = 2.5 \) min: The velocity increases from 45 to 75 km/h, indicating positive acceleration.
   - From \( t = 2.5 \) to \( t = 4 \) min: No change in velocity, so acceleration is 0.
   - From \( t = 4 \) to \( t = 4.5 \) min: The car decelerates to 0 km/h, showing negative acceleration.

2. **Distance Graph:**
   - Distance is the area under the velocity-time graph.
   - Calculate the area of each segment (rectangle or triangle) under the velocity curve to sketch the distance graph. Each segment represents the distance traveled over time.

### (b) Average Acceleration between \( t = 0 \) and \( t = 1 \) min
- **Formula**: \( a_{\text{avg}} = \frac{\Delta v}{\Delta t} \)
- Initial velocity \( v_0 = 0 \) km/h, final velocity \( v = 45 \) km/h, \( \Delta t = 1 \) min = \( \frac{1}{60} \) h.
\[
a_{\text{avg}} = \frac{45 - 0}{1/60} = 45 \times 60 = 2700 \text{ km/h}^2
\]

### (c) Average Acceleration between \( t = 2 \) and \( t = 3 \) min
- Initial velocity \( v_0 = 45 \) km/h, final velocity \( v = 75 \) km/h, \( \Delta t = 1 \) min = \( \frac{1}{60} \) h.
\[
a_{\text{avg}} = \frac{75 - 45}{1/60} = 30 \times 60 = 1800 \text{ km/h}^2
\]

Would you like a detailed sketch of the acceleration and distance graphs? Let me know if you have any questions!

### Related Questions:
1. How would the graph of acceleration differ for a more gradual acceleration phase?
2. How can the area under a velocity-time graph be used to find the exact distance traveled?
3. How do you determine instantaneous acceleration from a velocity-time graph?
4. What is the physical significance of negative acceleration on this graph?
5. How would the velocity-time graph change if the car started decelerating from \( t = 2 \) min instead of increasing speed?

### Tip:
Always remember that the slope of a velocity-time graph represents acceleration, and the area under it represents distance traveled.

The following graphic represents the marking of a speedometer of an automobile as a function of time. Sketch the corresponding graphs of the acceleration and the distance traveled by the car as a function of time. What is the average acceleration of the car between t = 0 and t = 1 min? What is the average acceleration of the car between t = 2 min and t = 3 min?

Analyze a velocity-time graph to sketch the corresponding acceleration and distance graphs. Learn to calculate average acceleration between t = 0 to t = 1 min and t = 2 to t = 3 min. This problem covers key concepts in kinematics, suitable for college-level physics.

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