KINEMATICS O13

Concept: Galilean Relativity. The river flows at a uniform velocity $\vec{u}$. Buoys released earlier have drifted further downstream than those released later. The final position of a buoy is $\vec{r}_{final} = \vec{r}_{initial} + \vec{u} \times (\text{drift time})$.

1. Setup and Release Order

Based on the arrows in the diagram, the buoys are released in the order A $\to$ B $\to$ C $\to$ D $\to$ E $\to$ F $\to$ G. Let the time interval between releases be $\tau$.

• Buoy A: Released at $t=0$. Total drift time = $6\tau$.
• Buoy G: Released at $t=6\tau$. Total drift time = $0$.

2. Determining the Flow

The problem states: “Buoy G falls on A”.

Let $\vec{p}_A$ and $\vec{p}_G$ be the positions where they were dropped (the static square pattern). In the static pattern, A is Top-Left and G is Top-Right.

Final Position of A: $\vec{R}_A = \vec{p}_A + \vec{u}(6\tau)$

Final Position of G: $\vec{R}_G = \vec{p}_G + \vec{u}(0) = \vec{p}_G$

Since they coincide ($\vec{R}_A = \vec{R}_G$):

The drift $6\tau\vec{u}$ exactly covers the distance from A to G. This means the river flows from left to right with a speed sufficient to push A exactly onto G’s spot.

3. Calculating Shifts for Other Buoys

Every buoy $k$ (where $k$ is the index 0 to 6) drifts by distance proportional to its time in water: $(6-k)\tau \vec{u}$.

Since total drift for A (index 0) is the vector $\vec{AG}$, the drift for any buoy is $\frac{6-k}{6} \vec{AG}$.

Let’s visualize the shifts relative to the static square grid:

• A (Index 0): Drifts 100% of distance A$\to$G. Lands on G.
• G (Index 6): Drifts 0%. Stays at G. (A and G are coincident at the top vertex).
• D (Index 3): Is physically in the middle horizontally. Drifts 50% of the distance. It ends up directly below the A/G point.
• C (Index 2): Starts at Bottom-Left. Drifts $\approx 66\%$. Ends up to the left of the centerline.
• E (Index 4): Starts at Bottom-Right. Drifts $\approx 33\%$. Ends up to the right of the centerline.