Solution: River-Boat Relative Motion
1. Analyzing the Frame of Reference
The most elegant way to solve this problem is to observe the motion from the River’s Frame of Reference (or equivalently, the frame of the floating buoy). In this frame:
- The water is stationary ($v_{river} = 0$).
- The buoy is stationary at the origin.
- The boats move with their relative velocities $v_1 = 4.0 \, \text{m/s}$ and $v_2 = 6.0 \, \text{m/s}$.
This simplifies the problem significantly because the drift of the river affects all objects (boats and buoy) equally.
2. Time Calculation
Since the boats move away from the buoy for a time $\tau$ and then turn back moving with the same speeds (relative to water), the time taken to return to the buoy must equal the time taken to move away.
Therefore, if $\tau$ is the time until they turn back:
- Total time of motion $T = \tau_{\text{out}} + \tau_{\text{return}} = 2\tau$.
We are given that the maximum separation between the boats is $200$ m. In the river frame, both boats move upstream initially relative to the ground, but relative to each other, they are separating.
The separation speed is the difference in their velocities:
$$v_{\text{sep}} = v_2 – v_1 = 6.0 – 4.0 = 2.0 \, \text{m/s}$$The separation distance $d$ after time $\tau$ is:
$$d = v_{\text{sep}} \times \tau$$ $$200 = 2.0 \times \tau \implies \tau = 100 \, \text{s}$$Total time elapsed: $T = 2\tau = 200 \, \text{s}$.
3. Distance in Ground Frame
We need to find the distance between the place where the buoy was dropped and the place where the faster boat passes the buoy again.
- Start Place: Location of buoy at $t=0$.
- End Place: Location of buoy at $t=200 \, \text{s}$.
Since the buoy drifts with the river velocity $v_r = 1.5 \, \text{m/s}$, the distance between these two places on the ground is simply the displacement of the buoy.
$$S = v_r \times T$$ $$S = 1.5 \, \text{m/s} \times 200 \, \text{s}$$ $$S = 300 \, \text{m}$$Correct Answer: (c) 300 m