Math Problem Statement

a radar gun was used to record the speed of a runner duriong the second of a race. Use simpsons rule to estimate the distance the runner covered in 5 seconds. The table has points (0,0), (0.5,3.15), (1,4.3), (1.5,5.45), (2,6.5), (2.5,7.45), (3, 8.95), (4,10.6), (4.5, 10.7), (5,10.7)

Solution

To estimate the distance the runner covered using Simpson's Rule, we treat the given points as measurements of speed at different times. The formula for Simpson's Rule, when applied to an odd number of intervals, is:

abf(x)dxh3[y0+4y1+2y2+4y3++2yn2+4yn1+yn]\int_{a}^{b} f(x) dx \approx \frac{h}{3} \left[ y_0 + 4y_1 + 2y_2 + 4y_3 + \dots + 2y_{n-2} + 4y_{n-1} + y_n \right]

Where:

  • h=banh = \frac{b-a}{n} is the width of each interval,
  • y0,y1,,yny_0, y_1, \dots, y_n are the function values (speeds in this case),
  • nn is the number of subintervals, which must be even.

Step-by-Step Solution:

  1. Data from the problem:

    • Time (seconds): t=[0,0.5,1,1.5,2,2.5,3,4,4.5,5]t = [0, 0.5, 1, 1.5, 2, 2.5, 3, 4, 4.5, 5]
    • Speed (meters per second): v=[0,3.15,4.3,5.45,6.5,7.45,8.95,10.6,10.7,10.7]v = [0, 3.15, 4.3, 5.45, 6.5, 7.45, 8.95, 10.6, 10.7, 10.7]
  2. Identify hh: The time intervals are evenly spaced by h=0.5h = 0.5 seconds.

  3. Apply Simpson's Rule formula: Since we have 9 intervals (an even number, which is suitable for Simpson's rule), we can proceed as follows:

Distanceh3[y0+4y1+2y2+4y3+2y4+4y5+2y6+4y7+2y8+y9]\text{Distance} \approx \frac{h}{3} \left[ y_0 + 4y_1 + 2y_2 + 4y_3 + 2y_4 + 4y_5 + 2y_6 + 4y_7 + 2y_8 + y_9 \right]

Substituting the values from the table:

Distance0.53[0+4(3.15)+2(4.3)+4(5.45)+2(6.5)+4(7.45)+2(8.95)+4(10.6)+2(10.7)+10.7]\text{Distance} \approx \frac{0.5}{3} \left[ 0 + 4(3.15) + 2(4.3) + 4(5.45) + 2(6.5) + 4(7.45) + 2(8.95) + 4(10.6) + 2(10.7) + 10.7 \right]

  1. Calculate: Simplifying inside the brackets:

=0.5×13×[0+12.6+8.6+21.8+13+29.8+17.9+42.4+21.4+10.7]= 0.5 \times \frac{1}{3} \times \left[ 0 + 12.6 + 8.6 + 21.8 + 13 + 29.8 + 17.9 + 42.4 + 21.4 + 10.7 \right] =0.5×13×[177.5]= 0.5 \times \frac{1}{3} \times \left[ 177.5 \right] =0.5×59.167= 0.5 \times 59.167 =29.5835 meters= 29.5835 \text{ meters}

Conclusion:

The estimated distance the runner covered in 5 seconds, using Simpson's Rule, is approximately 29.58 meters.

Would you like to explore any part of this in more detail or have any questions?

Here are 5 related questions to think about:

  1. How does Simpson's rule compare with other numerical integration methods, like the trapezoidal rule?
  2. What would happen if the time intervals were not evenly spaced?
  3. Can Simpson's rule be applied to estimate the area under any curve?
  4. How would errors accumulate if the given data were less accurate?
  5. Can you apply Simpson's rule to other real-life problems, like estimating total fuel consumption from speed data?

Tip: Simpson’s rule works best when the data approximates a smooth curve; otherwise, more intervals might be necessary to improve accuracy.

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Math Problem Analysis

Mathematical Concepts

Numerical Integration
Simpson's Rule
Distance Estimation

Formulas

Simpson's Rule: \( \int_{a}^{b} f(x) dx \approx \frac{h}{3} \left[ y_0 + 4y_1 + 2y_2 + \dots + y_n \right] \)
Distance Formula: Distance = Speed × Time

Theorems

Simpson's Rule for numerical integration

Suitable Grade Level

Grades 10-12