Question

Racing cars driven by Chris and Kelly are side by side at the start of a race. The table shows the velocities of each car (in miles per hour) during the first ten seconds of the race. Use Simpson's Rule to estimate how much farther Kelly travels than Chris does during the first ten seconds.

Short answer

Kelley travels \(80.67{\rm{ft}}\) more than Chris during first 10 seconds.

Step-by-step solution

The general expression for Simpson’s rule

Simpson's rule:

\(\int_a^b f (x)dx \approx \frac{{\Delta t}}{3}\left( {{y_0} + 4{y_1} + 2{y_2} + 4{y_3} + 2{y_4} + \cdots + 4{y_{n - 1}} + {y_n}} \right)\)

where\(\Delta t = \frac{{b - a}}{n}\)

Distance between Kelly and Chris during first 10 seconds.

Find the width of the subinterval \((\Delta t)\) as follows.

Consider the number of subintervals as 5 .

Substitute 10 for n, 10 for \(b\) and 0 for \(a\) in \(\Delta x = \frac{{b - a}}{n}\) as follows.

\(\Delta t = \frac{{10 - 0}}{{10}} = 1\)

Evaluate the distance between Kelly and Chris as follows.

\(\begin{aligned}{l}\int\limits_0^{10} {\left( {{v_{\rm{K}}} - {v_{\rm{C}}}} \right)dt} \\ \approx \frac{1}{3}(0 + 4(2) + 2(5) + 4(6) + 2(7) + 4(9) + 2(11) + 4(11) + 2(12) + 4(12) + 12)\\ \approx \frac{1}{3}(0 + 8 + 10 + 24 + 14 + 36 + 22 + 44 + 24 + 48 + 12)\\ \approx \frac{1}{3}(242)\end{aligned}\)

Solve the steps further

\(\begin{aligned}{l} \approx 80.666\\ \approx 80.67\end{aligned}\)

Therefore, Kelley travels \(80.67{\rm{ft}}\) more than Chris during first 10 seconds.