Question

A radar gun was used to record the speed of a runner during the first 5 seconds of a race (see the table). Use simpson's rule to estimate the distance the runner covered during those 5 seconds. (round your answer to three decimal places. ).

Answer 1

The sum of the product of the times and velocity is given by the definite

integral of the function of velocity which is given by Simpson's rule.

Response:

• The distance the runner covered during the 5 seconds is approximately 44.772 m

What is the Simpson's rule, and how can it be used to estimate the distance?

The Simpson's rule is used for to approximate a definite integral

The possible table of values of the velocity and time obtained from a similar question is presented as follows;

$\begin{array}{|c|c|c}t(s)&v(m/s)&\\0&0\\0.5&4.67\\1.0&7.32&\\1.5&8.84&\\2.0&9.73&\\2.5&10.24&\\3.0&10.57&\\3.5&10.63&\\4.0&10.76&\\4.5&10.87&\\5.0&10.87&\end{array}\right]$

Distance = Velocity × Time

$The \ distance \ is \ given \ by \ the \ integral, \ \mathbf{\displaystyle \int\limits^a_b {f(t)} \, dt}$

Where;

f(t) = v

dt = Small duration in time

According to Simpson's rule, we have;

$\displaystyle \int\limits^a_b {f(x)} \, dx = \mathbf{\frac{\Delta x}{3} \left(f(x_0) + 4 \cdot f(x_1) + 2\cdot f(x_3) + ... + 2\cdot f(x_{n - 2}) + 4 \cdot f(x_{n-1}) + f(x_n) \right)}$

Δx = $\mathbf{\dfrac{b - a}{n}}$

n = 10

Which gives;

$\dfrac{5.0 - 0}{10} = \dfrac{1}{2} = 0.5$

$d = \displaystyle \int\limits^a_b {f(t)} \, dt = \mathbf{ \frac{0.5}{3} \left(f(t_0) + 4 \cdot f(t_1) + 2\cdot f(t_3) + ... + 2\cdot f(t_{10 - 2}) + 4 \cdot f(t_{10-1}) + f(x_{10}) \right)}$

Therefore;

d = 0.5÷3×(0 + 4 × 4.67 + 2 × 7.32 + 4 × 8.84 + 2 × 9.73 + 4 × 10.24 + 2 × 10.57 + 4 × 10.63 + 2 × 10.76 + 4 × 10.87 + 10.87) ≈ 44.772

• The estimate of the distance covered during the 5 seconds, d ≈ 44.772 m

Learn more about the definite integrals here:

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