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### Given:
- The equation of motion of the mass is: 
  \[
  s(t) = 2\cos(t) + 3\sin(t)
  \]
  where \( s(t) \) is the position of the mass (in centimeters), and \( t \) is the time (in seconds).
- Positive direction for \( s(t) \) is downward.

### Part (a): Find the velocity function.
The velocity function \( v(t) \) is the derivative of the position function \( s(t) \) with respect to \( t \). 

\[
v(t) = \frac{ds(t)}{dt}
\]
Taking the derivative of \( s(t) = 2\cos(t) + 3\sin(t) \):

\[
v(t) = -2\sin(t) + 3\cos(t)
\]

### Part (b): When does the mass pass through the equilibrium position for the first time?
The equilibrium position is when \( s(t) = 0 \). We need to solve:

\[
s(t) = 2\cos(t) + 3\sin(t) = 0
\]

Dividing both sides by 2 gives:

\[
\cos(t) + \frac{3}{2}\sin(t) = 0
\]

We can solve this for \( t \).

### Part (c): How far from its equilibrium position does the mass travel?
The amplitude of the oscillation will give the maximum displacement of the mass from its equilibrium position. This is determined by finding the maximum value of \( s(t) \).

Let me now solve the parts (b) and (c).### Part (b): 
The mass passes through the equilibrium position for the first time at \( t = -\tan^{-1}(2/3) \). In radians, this simplifies to approximately \( t = -0.59 \) seconds. However, since time cannot be negative, the correct first positive time can be adjusted by adding \( \pi \), giving us \( t \approx 2.55 \) seconds (since we are working in periodic trigonometric functions).

### Part (c): 
The amplitude of the oscillation is the distance from the equilibrium position, which is approximately \( 3.61 \) cm.

### Summary of Answers:
- **Velocity function:** \( v(t) = -2\sin(t) + 3\cos(t) \)
- **First time crossing equilibrium position:** \( t \approx 2.55 \) seconds
- **Maximum distance from equilibrium:** \( d \approx 3.61 \) cm

Would you like further clarification on any part of this, or more details?

Here are 5 questions related to this:
1. What is the period of the oscillation for this system?
2. How do we determine the general solution for equilibrium crossings?
3. What happens to the velocity function at the equilibrium point?
4. What if the initial displacement was different? How would that change the amplitude?
5. How can damping affect the equation of motion and the amplitude?

Tip: When solving oscillatory problems, it's essential to understand the interplay between position, velocity, and acceleration functions.

An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is pulled downward and then released, it vibrates vertically. The equation of motion is s(t) = 2cos(t) + 3sin(t), where s is measured in centimeters and t is in seconds. (a) Find the velocity function. (b) When does the mass pass through the equilibrium position for the first time? (c) How far from its equilibrium position does the mass travel?

Learn how to solve for the velocity function, equilibrium time, and maximum displacement of a mass hanging from an elastic band vibrating vertically using the equation of motion s(t) = 2cos(t) + 3sin(t). Step-by-step differentiation and trigonometric problem-solving are provided. Suitable for Grades 11-12.

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