Joshua's Merry-go-round

We are given that the merry-go-round has a radius of $4$ meters. Joshua initially stands $2$ meters away from the nearest point on his merry-go-round, and he experiences a total displacement of $\dfrac{548\pi-3}{\pi}$ meters after walking through point $B$ to point $C$ on the merry-go-ground and standing there for exactly 3 minutes while it was being operated before walking back to his initial position. We are also given that the merry-go-round has a rotational frequency of $\pi \dfrac{\text{radians}}{\text{second}}$ when he stands on it.

From this, we can set up a linear equation to solve for the distance between $B$ and $C$ . If we let the distance between $B$ and $C$ be $s$ , the distance from the center of the merry-go-round to point $C$ is $4-s$ .

Hence, Joshua experienced a displacement of $2\pi(4-s) \dfrac{\text{meters}}{\text{revs}} \times 90 \text{ revs} = 180\pi(4-s) \text{ meters}$ from the rotations of his merry-go-round. In his path, Joshua walked from $B$ to $C$ and back, covering $2s$ meters, and from $A$ to $B$ and back, covering $4$ meters.

Since his total displacement is $\dfrac{548\pi-3}{\pi}$ , we see that $180\pi(4-s)+2s+4 = \dfrac{548\pi-3}{\pi} \\ \implies (2-180\pi)s = \dfrac{548\pi-3}{\pi}-4-720\pi \implies s = \dfrac{\dfrac{548\pi-3}{\pi}-\dfrac{4\pi}{\pi}-720\pi}{2-180\pi} = \dfrac{\dfrac{544\pi-3}{\pi}-720\pi}{2-180\pi} \\ \therefore s \approx 3.05047 \implies \lfloor {s} \rfloor = \space \boxed{3}$ .