1. Identify the given information and the goal:
- We have a unit circle with points $A$ and $B$ such that the chord $AB$ has length 1.
- We need to find the point $C$ on the longer arc $AB$ that maximizes the perimeter of triangle $ABC$.

2. **Calculate the central angle $\theta$ subtended by chord $AB$:**
- Since $AB$ is a chord of length 1 in a unit circle, we can use the chord length formula:
$$AB = 2 \sin\left(\frac{\theta}{2}\right)$$
Given $AB = 1$, we have:
$$1 = 2 \sin\left(\frac{\theta}{2}\right) \implies \sin\left(\frac{\theta}{2}\right) = \frac{1}{2} \implies \frac{\theta}{2} = 30^\circ \implies \theta = 60^\circ$$

3. **Express the perimeter of $\triangle ABC$:**
- Let $\alpha$ be the angle subtended by arc $AC$ at the center of the circle.
- The perimeter of $\triangle ABC$ is given by:
$$P = AB + AC + BC$$
- Since $AB = 1$, we need to find $AC$ and $BC$.

4. **Calculate $AC$ and $BC$:**
- The lengths $AC$ and $BC$ are the chord lengths corresponding to angles $\alpha$ and $120^\circ - \alpha$ respectively.
- Using the chord length formula:
$$AC = 2 \sin\left(\frac{\alpha}{2}\right)$$
$$BC = 2 \sin\left(\frac{120^\circ - \alpha}{2}\right)$$

5. Formulate the expression for the perimeter:
- The perimeter $P$ is:
$$P = 1 + 2 \sin\left(\frac{\alpha}{2}\right) + 2 \sin\left(\frac{120^\circ - \alpha}{2}\right)$$

6. Simplify the expression:
- Let $x = \frac{\alpha}{2}$. Then the perimeter becomes:
$$P = 1 + 2 \sin(x) + 2 \sin(60^\circ - x)$$
- Using the angle subtraction formula for sine:
$$\sin(60^\circ - x) = \sin 60^\circ \cos x - \cos 60^\circ \sin x = \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x$$
- Therefore:
$$P = 1 + 2 \sin(x) + 2 \left( \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x \right) = 1 + 2 \sin(x) + \sqrt{3} \cos x - \sin(x)$$
$$P = 1 + \sin(x) + \sqrt{3} \cos x$$

7. Maximize the expression:
- To maximize $P = 1 + \sin(x) + \sqrt{3} \cos x$, we use the Cauchy-Schwarz inequality:
$$(\sin x + \sqrt{3} \cos x)^2 \leq (1^2 + (\sqrt{3})^2)(\sin^2 x + \cos^2 x) = 4$$
$$\sin x + \sqrt{3} \cos x \leq 2$$
- Therefore, the maximum value of $\sin x + \sqrt{3} \cos x$ is 2, and the maximum perimeter is:
$$P = 1 + 2 = 3$$

The final answer is $\boxed{3}$.