The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s = 2sin(𝜋t) + 2cos(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]
  cm/s

(ii)    [1, 1.1]
  cm/s

(iii)    [1, 1.01]
  cm/s

(iv)    [1, 1.001]
  cm/s

(b) Estimate the instantaneous velocity of the particle when t = 1.
  cm/s

Given the displacement function \( s(t) = 2\sin(\pi t) + 2\cos(\pi t) \), we need to calculate the average velocity over various time intervals and then estimate the instantaneous velocity at \( t = 1 \).

### (a) Average Velocity Calculation
The average velocity over a time interval \([t_1, t_2]\) is given by:
\[
\text{Average velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}
\]
where \( s(t) \) is the displacement function.

#### (i) For the interval \([1, 2]\):
\[
s(1) = 2\sin(\pi \times 1) + 2\cos(\pi \times 1) = 2\sin(\pi) + 2\cos(\pi) = 2(0) + 2(-1) = -2 \text{ cm}
\]
\[
s(2) = 2\sin(\pi \times 2) + 2\cos(\pi \times 2) = 2\sin(2\pi) + 2\cos(2\pi) = 2(0) + 2(1) = 2 \text{ cm}
\]
\[
\text{Average velocity} = \frac{2 - (-2)}{2 - 1} = \frac{4}{1} = 4 \text{ cm/s}
\]

#### (ii) For the interval \([1, 1.1]\):
\[
s(1.1) = 2\sin(\pi \times 1.1) + 2\cos(\pi \times 1.1) = 2\sin(1.1\pi) + 2\cos(1.1\pi)
\]
\[
s(1.1) = 2(-0.309) + 2(0.951) \approx -0.618 + 1.902 = 1.284 \text{ cm}
\]
\[
\text{Average velocity} = \frac{1.284 - (-2)}{1.1 - 1} = \frac{3.284}{0.1} = 32.84 \text{ cm/s}
\]

#### (iii) For the interval \([1, 1.01]\):
\[
s(1.01) = 2\sin(\pi \times 1.01) + 2\cos(\pi \times 1.01) = 2\sin(1.01\pi) + 2\cos(1.01\pi)
\]
\[
s(1.01) = 2(-0.0314) + 2(0.9995) \approx -0.0628 + 1.999 = 1.9362 \text{ cm}
\]
\[
\text{Average velocity} = \frac{1.9362 - (-2)}{1.01 - 1} = \frac{3.9362}{0.01} = 393.62 \text{ cm/s}
\]

#### (iv) For the interval \([1, 1.001]\):
\[
s(1.001) = 2\sin(\pi \times 1.001) + 2\cos(\pi \times 1.001) = 2\sin(1.001\pi) + 2\cos(1.001\pi)
\]
\[
s(1.001) = 2(-0.00314) + 2(0.999995) \approx -0.00628 + 1.99999 = 1.99371 \text{ cm}
\]
\[
\text{Average velocity} = \frac{1.99371 - (-2)}{1.001 - 1} = \frac{3.99371}{0.001} = 3993.71 \text{ cm/s}
\]

### (b) Instantaneous Velocity at \( t = 1 \)
The instantaneous velocity at \( t = 1 \) can be estimated by finding the derivative of \( s(t) \) with respect to \( t \) and then evaluating it at \( t = 1 \).

First, find \( s'(t) \):
\[
s(t) = 2\sin(\pi t) + 2\cos(\pi t)
\]
\[
s'(t) = 2\pi\cos(\pi t) - 2\pi\sin(\pi t)
\]

At \( t = 1 \):
\[
s'(1) = 2\pi\cos(\pi) - 2\pi\sin(\pi) = 2\pi(-1) - 2\pi(0) = -2\pi \text{ cm/s} \approx -6.28 \text{ cm/s}
\]

### Final Answers:
(a) Average velocity:
- (i) [1, 2]: \( 4 \) cm/s
- (ii) [1, 1.1]: \( 32.84 \) cm/s
- (iii) [1, 1.01]: \( 393.62 \) cm/s
- (iv) [1, 1.001]: \( 3993.71 \) cm/s

(b) Instantaneous velocity at \( t = 1 \): \( -6.28 \) cm/s

Would you like more details or have any questions?

### Related Questions:
1. How do you calculate average velocity for a different function?
2. What are the physical interpretations of positive and negative instantaneous velocities?
3. How would the results change if the displacement function included a quadratic term?
4. What happens to the velocity as the time interval decreases further?
5. How can you use limits to find the instantaneous velocity more rigorously?

### Tip:
Always ensure to keep track of units when performing calculations, especially when dealing with physical quantities like displacement, velocity, and time.

The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s = 2sin(𝜋t) + 2cos(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]  cm/s
(ii)    [1, 1.1]  cm/s
(iii)    [1, 1.01]  cm/s
(iv)    [1, 1.001]  cm/s
(b) Estimate the instantaneous velocity of the particle when t = 1.  cm/s

Calculate the average velocity of a particle over various time intervals and estimate the instantaneous velocity using the displacement function s = 2sin(πt) + 2cos(πt). This problem covers trigonometry and calculus concepts, suitable for students in Grades 11-12.

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