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To estimate the distance traveled by the train in the first 20 seconds using four equal strips (5 seconds each), we can apply the method of rectangles under the curve.

The graph shows velocity as a function of time, indicating a nonlinear increase in velocity, suggesting that the train is accelerating.

### Step-by-step Calculation:

1. **Divide the 20 seconds into 4 strips of 5 seconds each.**
2. **Estimate the velocity at each 5-second interval using the graph:**
   - At \( t = 0 \) s, \( v = 0 \) m/s
   - At \( t = 5 \) s, \( v \approx 1.5 \) m/s
   - At \( t = 10 \) s, \( v \approx 4 \) m/s
   - At \( t = 15 \) s, \( v \approx 9 \) m/s
   - At \( t = 20 \) s, \( v \approx 17 \) m/s

3. **Calculate the distance for each interval using the midpoint velocity (approximate height of each rectangle):**
   - \( 0 \) to \( 5 \) s: \( \text{Distance} \approx 1.5 \, \text{m/s} \times 5 \, \text{s} = 7.5 \, \text{m} \)
   - \( 5 \) to \( 10 \) s: \( \text{Distance} \approx 4 \, \text{m/s} \times 5 \, \text{s} = 20 \, \text{m} \)
   - \( 10 \) to \( 15 \) s: \( \text{Distance} \approx 9 \, \text{m/s} \times 5 \, \text{s} = 45 \, \text{m} \)
   - \( 15 \) to \( 20 \) s: \( \text{Distance} \approx 17 \, \text{m/s} \times 5 \, \text{s} = 85 \, \text{m} \)

4. **Add the distances from each interval:**
   - \( \text{Total Distance} \approx 7.5 \, \text{m} + 20 \, \text{m} + 45 \, \text{m} + 85 \, \text{m} = 157.5 \, \text{m} \)

### Answer to Part (a)
The estimated distance the train traveled in the first 20 seconds is approximately 157.5 meters.

### Part (b): Underestimate or Overestimate?
Using rectangles in this way (by picking the velocity at the beginning of each interval) generally results in an underestimate when the function is increasing, as the actual velocity throughout each interval is higher than at its start. Since the graph shows an accelerating train, the method of using the starting velocity of each interval likely underestimates the actual distance traveled.

Would you like more details or have any questions on this?

#### Related Questions
1. How would the distance estimate change if we used the end velocities of each interval instead?
2. What would be the distance estimate if we used the trapezoidal rule for this calculation?
3. How would calculus provide a more accurate answer to this problem?
4. What is the significance of the train's acceleration in this context?
5. How would the results differ if the velocity-time graph showed a constant velocity?

#### Tip
When estimating areas under a curve, remember that using different methods (rectangles, trapezoids, parabolic segments) can yield different results, with some methods providing more accurate estimates than others depending on the function's behavior.

Use four strips of equal width to estimate the distance the train traveled in the first 20 seconds. Is your answer an underestimate or an overestimate?

Learn how to estimate the distance traveled by a train using four equal strips under a velocity-time graph. The problem involves applying the rectangle approximation method for numerical integration. Suitable for high school students in Grades 10-12.

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