The speed of a motor launch with respect to still water in a stream is `8 ms^(-1)` while water current's speed is `3 m s^(-1)`. When the launch began travelling upstream, a float was dropped from it. After travelling a distance of `4.8 km` upstream, the launch turned back and caught up with the float. What is the total time which elapsed during the process?

Explanation

To solve the problem step by step, we will break it down into manageable parts.

### Step 1: Determine the effective speed of the motor launch upstream
The speed of the motor launch with respect to still water is \(8 \, \text{m/s}\) and the speed of the water current is \(3 \, \text{m/s}\). When the launch is traveling upstream, its effective speed is reduced by the speed of the current.

\[ \text{Effective speed upstream} = \text{Speed of launch} - \text{Speed of current} = 8 \, \text{m/s} - 3 \, \text{m/s} = 5 \, \text{m/s} \]
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