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Let's solve this step by step.

### Given:
The position function of a particle is:
\[
s(t) = \frac{t}{1 + t^2}.
\]

---

### Part (a): Find \(v(t)\) and \(a(t)\)
1. **Velocity (\(v(t)\))**: The velocity is the derivative of the position function:
   \[
   v(t) = \frac{ds(t)}{dt}.
   \]
   Using the quotient rule:
   \[
   \frac{d}{dt} \left( \frac{t}{1 + t^2} \right) = \frac{(1 + t^2)(1) - t(2t)}{(1 + t^2)^2}.
   \]
   Simplify:
   \[
   v(t) = \frac{1 + t^2 - 2t^2}{(1 + t^2)^2} = \frac{1 - t^2}{(1 + t^2)^2}.
   \]

   So,
   \[
   v(t) = \frac{1 - t^2}{(1 + t^2)^2}.
   \]

2. **Acceleration (\(a(t)\))**: The acceleration is the derivative of the velocity:
   \[
   a(t) = \frac{dv(t)}{dt}.
   \]
   Using the quotient rule again:
   \[
   \frac{d}{dt} \left( \frac{1 - t^2}{(1 + t^2)^2} \right) = \frac{(1 + t^2)^2 \cdot (-2t) - (1 - t^2) \cdot 2(1 + t^2)(2t)}{(1 + t^2)^4}.
   \]
   Simplify the numerator step-by-step:
   \[
   \text{Numerator} = -2t(1 + t^2)^2 - 2t(1 - t^2)(2)(1 + t^2).
   \]
   Factor out \( -2t(1 + t^2) \):
   \[
   \text{Numerator} = -2t(1 + t^2)\left[(1 + t^2) + 2(1 - t^2)\right].
   \]
   Simplify the bracket:
   \[
   (1 + t^2) + 2(1 - t^2) = 1 + t^2 + 2 - 2t^2 = 3 - t^2.
   \]
   So:
   \[
   \text{Numerator} = -2t(1 + t^2)(3 - t^2).
   \]
   The acceleration becomes:
   \[
   a(t) = \frac{-2t(1 + t^2)(3 - t^2)}{(1 + t^2)^4}.
   \]
   Cancel one \((1 + t^2)\):
   \[
   a(t) = \frac{-2t(3 - t^2)}{(1 + t^2)^3}.
   \]

---

### Part (b): Determine intervals where the object is slowing down or speeding up
- The particle is:
  - **Speeding up** when \(v(t)\) and \(a(t)\) have the **same sign**.
  - **Slowing down** when \(v(t)\) and \(a(t)\) have **opposite signs**.

#### Analyze \(v(t)\):
\[
v(t) = \frac{1 - t^2}{(1 + t^2)^2}.
\]
- \(v(t) > 0\) when \(1 - t^2 > 0 \implies t \in (0, 1)\).
- \(v(t) < 0\) when \(1 - t^2 < 0 \implies t \in (1, \infty)\).

#### Analyze \(a(t)\):
\[
a(t) = \frac{-2t(3 - t^2)}{(1 + t^2)^3}.
\]
- \(a(t) > 0\) when:
  - \(-2t < 0\) (so \(t > 0\)) **and** \(3 - t^2 > 0 \implies t \in (0, \sqrt{3})\).
- \(a(t) < 0\) when:
  - \(-2t < 0\) (so \(t > 0\)) **and** \(3 - t^2 < 0 \implies t \in (\sqrt{3}, \infty)\).

#### Combine signs of \(v(t)\) and \(a(t)\):
1. **Speeding up** (\(v(t)\) and \(a(t)\) have the same sign):
   - \(t \in (0, 1)\) (both \(v(t) > 0\) and \(a(t) > 0\)).
   - \(t \in (\sqrt{3}, \infty)\) (both \(v(t) < 0\) and \(a(t) < 0\)).

2. **Slowing down** (\(v(t)\) and \(a(t)\) have opposite signs):
   - \(t \in (1, \sqrt{3})\) (\(v(t) < 0\), \(a(t) > 0\)).

---

### Final Answers:
(a)  
\[
v(t) = \frac{1 - t^2}{(1 + t^2)^2}, \quad a(t) = \frac{-2t(3 - t^2)}{(1 + t^2)^3}.
\]

(b)  
- Slowing down: \(t \in (1, \sqrt{3})\).  
- Speeding up: \(t \in (0, 1) \cup (\sqrt{3}, \infty)\).

---

Would you like a deeper explanation of any part?  

Here are 5 related questions you could explore:
1. How does the quotient rule apply to more complex functions?
2. Why do we check the signs of velocity and acceleration for speeding up/slowing down?
3. How do inflection points of \(s(t)\) relate to acceleration?
4. What happens when \(t = 1\) or \(t = \sqrt{3}\)?
5. Can this method be extended to 2D or 3D motion?

**Tip:** Always double-check the domains of your functions to ensure intervals make physical sense.

Find the velocity and acceleration functions of a particle with position function s(t) = t / (1 + t^2), and determine intervals where the particle is speeding up or slowing down.

This problem covers the calculation of velocity and acceleration for a particle with position function s(t) = t / (1 + t^2), using derivatives and the quotient rule. The solution also analyzes intervals where the particle is speeding up or slowing down, based on the signs of velocity and acceleration. Suitable for students in Grades 11-12 studying calculus.

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