The average velocity of a particle over a time interval $[t_1, t_2]$ is given by the formula:

$v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$

where $s(t) = 5 \sin (\pi t) + 2 \cos (\pi t)$.

Step 1: Compute $s(1)$

$s(1) = 5 \sin (\pi \cdot 1) + 2 \cos (\pi \cdot 1)$

$= 5 \sin \pi + 2 \cos \pi$

$= 5(0) + 2(-1) = -2$

Step 2: Compute $s(1.01)$

$s(1.01) = 5 \sin (\pi \cdot 1.01) + 2 \cos (\pi \cdot 1.01)$

Approximating values:

$\sin (\pi \cdot 1.01) = \sin (\pi + 0.01\pi) = -\sin (0.01\pi) \approx -0.0314$

$\cos (\pi \cdot 1.01) = \cos (\pi + 0.01\pi) = -\cos (0.01\pi) \approx -0.9995$

$s(1.01) = 5(-0.0314) + 2(-0.9995)$

$= -0.157 - 1.999 = -2.156$

Step 3: Compute the Average Velocity

$v_{\text{avg}} = \frac{s(1.01) - s(1)}{1.01 - 1}$

$= \frac{-2.156 - (-2)}{0.01}$

$= \frac{-2.156 + 2}{0.01}$

$= \frac{-0.156}{0.01} = -15.6$

Final Answer:

$v_{\text{avg}} \approx -15.60 \text{ cm/s}$

Would you like a more detailed explanation? Here are some related questions:

1. How do we approximate sine and cosine for small angles?
2. Why do we use the difference quotient to calculate average velocity?
3. What is the significance of a negative average velocity?
4. How does average velocity differ from instantaneous velocity?
5. How would the result change for a different interval, such as [1, 1.1]?

Tip: The negative sign of the average velocity means the particle is moving in the negative direction over the given time interval.