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.