The displacement (in centimeters) of a particle s moving back and forth along a straight line is given by the equation
s = 4 sin(𝜋t) + 2 cos(𝜋t),
where t is measured in seconds. (Round your answers to two decimal places.)
(a)
Find the average velocity during each time period.
(i)
[1, 2]
4

Correct: Your answer is correct.
cm/s
(ii)
[1, 1.1]
-11.38

Correct: Your answer is correct.
cm/s
(iii)
[1, 1.01]
-12.56

Incorrect: Your answer is incorrect.
cm/s
(iv)
[1, 1.001]
-12.57

Incorrect: Your answer is incorrect.
cm/s
(b)
Estimate the instantaneous velocity of the particle when
t = 1.
-12.57

Correct: Your answer is correct.
cm/s

Let's address the issue you're encountering with parts (iii) and (iv) of your problem.

### (a) Average Velocity Calculation

The average velocity \( \bar{v} \) over a time interval \([t_1, t_2]\) is given by:

\[
\bar{v} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}
\]

For the displacement function \( s(t) = 4 \sin(\pi t) + 2 \cos(\pi t) \), we need to evaluate the displacement at the endpoints of the intervals and use the formula above.

#### (iii) Time Interval \([1, 1.01]\)
Let's calculate the average velocity for this interval.

1. **Calculate \( s(1) \):**
   \[
   s(1) = 4 \sin(\pi \times 1) + 2 \cos(\pi \times 1) = 4 \times 0 + 2 \times (-1) = -2 \, \text{cm}
   \]

2. **Calculate \( s(1.01) \):**
   \[
   s(1.01) = 4 \sin(\pi \times 1.01) + 2 \cos(\pi \times 1.01)
   \]

   We'll compute this value.

3. **Calculate Average Velocity:**
   \[
   \bar{v} = \frac{s(1.01) - s(1)}{1.01 - 1}
   \]

#### (iv) Time Interval \([1, 1.001]\)
Similarly, calculate the average velocity for this interval.

1. **Calculate \( s(1.001) \):**
   \[
   s(1.001) = 4 \sin(\pi \times 1.001) + 2 \cos(\pi \times 1.001)
   \]

2. **Calculate Average Velocity:**
   \[
   \bar{v} = \frac{s(1.001) - s(1)}{1.001 - 1}
   \]

Finally, we'll estimate the instantaneous velocity at \( t = 1 \) using the concept of limits. Let's compute these values.Here are the results:

1. **Displacement at \( t = 1 \):** \( s(1) \approx -2 \, \text{cm} \)
2. **Displacement at \( t = 1.01 \):** \( s(1.01) \approx -2.1247 \, \text{cm} \)
3. **Displacement at \( t = 1.001 \):** \( s(1.001) \approx -2.0126 \, \text{cm} \)

### Average Velocities:
- **(iii) For the interval [1, 1.01]:**
  \[
  \bar{v} \approx -12.47 \, \text{cm/s}
  \]

- **(iv) For the interval [1, 1.001]:**
  \[
  \bar{v} \approx -12.56 \, \text{cm/s}
  \]

These results suggest that the average velocity over smaller intervals is approaching a value near \(-12.57 \, \text{cm/s}\), which matches the correct instantaneous velocity estimate at \( t = 1 \).

If you need further clarification or additional details, feel free to ask!

### Related Questions:
1. How does the displacement function affect velocity?
2. What is the relationship between instantaneous and average velocity?
3. How would acceleration be calculated from this function?
4. What does it mean for velocity to be negative?
5. How would the velocity change if the function were altered?

### Tip:
For instantaneous velocity, the smaller the time interval used, the closer the average velocity will approximate the actual instantaneous velocity.

Learn how to calculate average and instantaneous velocities for a particle moving along a straight line using trigonometric functions. Solve specific time interval problems with detailed step-by-step solutions. Suitable for college-level students studying calculus and trigonometry.

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