Questions: A radar gun was used to record the speed of a runner during the first 5 seconds of a race (see table). Use Simpson's rule to estimate the distance the runner covered during those 5 seconds. t(s) 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 v(ft / s) 0 3.05 3.25 7.1 7.7 9.25 9.55 10.35 10.7 10.85 10.85

Solution Steps

We have the following data for time \( t \) (in seconds) and velocity \( v \) (in ft/s):

\[ t = [0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5] \] \[ v = [0, 3.05, 3.25, 7.1, 7.7, 9.25, 9.55, 10.35, 10.7, 10.85, 10.85] \]

The number of intervals \( n \) is calculated as:

\[ n = \text{len}(t) - 1 = 11 - 1 = 10 \]

The step size \( h \) is given by:

\[ h = \frac{t[-1] - t[0]}{n} = \frac{5 - 0}{10} = 0.5 \]

Using Simpson's rule, the estimated distance \( D \) is calculated as follows:

\[ D = v[0] + v[-1] + \sum_{i=1}^{n-1} \left(4 \cdot v[i] \text{ if } i \text{ is odd} + 2 \cdot v[i] \text{ if } i \text{ is even}\right) \]

Calculating the contributions:

• For \( i = 1 \): \( 4 \cdot v[1] = 4 \cdot 3.05 = 12.2 \)
• For \( i = 2 \): \( 2 \cdot v[2] = 2 \cdot 3.25 = 6.5 \)
• For \( i = 3 \): \( 4 \cdot v[3] = 4 \cdot 7.1 = 28.4 \)
• For \( i = 4 \): \( 2 \cdot v[4] = 2 \cdot 7.7 = 15.4 \)
• For \( i = 5 \): \( 4 \cdot v[5] = 4 \cdot 9.25 = 37.0 \)
• For \( i = 6 \): \( 2 \cdot v[6] = 2 \cdot 9.55 = 19.1 \)
• For \( i = 7 \): \( 4 \cdot v[7] = 4 \cdot 10.35 = 41.4 \)
• For \( i = 8 \): \( 2 \cdot v[8] = 2 \cdot 10.7 = 21.4 \)
• For \( i = 9 \): \( 4 \cdot v[9] = 4 \cdot 10.85 = 43.4 \)

Summing these contributions:

\[ D = 0 + 10.85 + (12.2 + 6.5 + 28.4 + 15.4 + 37.0 + 19.1 + 41.4 + 21.4 + 43.4) = 0 + 10.85 + 224.3 = 235.15 \]

Now, we multiply by \( \frac{h}{3} \):

\[ D = \frac{h}{3} \cdot 235.15 = \frac{0.5}{3} \cdot 235.15 = \frac{235.15}{6} \approx 39.275 \]

Final Answer