Given:
The distance to the fishing spot, S=5 miles.
The time taken to reach the fishing spot travelling against the current, t=5/12 h.
The time taken to reach the fishing spot travelling with the current , T=1/4 h.
Now, the speed of the boat when it is travelling in the direction of the current or the downstream speed is,
![\begin{gathered} D=(S)/(T) \\ =(5)/((1)/(4)) \\ =5*4 \\ =20\text{ miles/h} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/w8dgtm1zoejf63jea3rdqisw0ohh0patqo.png)
The speed of the boat when it is travelling in the against the direction of the current or the upstream speed is,
![\begin{gathered} U=(D)/(t) \\ =(5)/((5)/(12)) \\ =5*(12)/(5) \\ =12\text{ miles/h} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/r2587651onwd16v56ruvktlg4z5tenhqwg.png)
Let b be the speed of the boat in still water and w be the speed of the current.
Hence, the speed of the boat in still water can be calculated as,
![\begin{gathered} b=(1)/(2)(D+U) \\ =(1)/(2)(20+12) \\ =(1)/(2)*32 \\ =16\text{ miles/h} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2e4g1h4jq7j22iib2us76ik0ulxjen5zoy.png)
The speed of the current can be calculated as,
![\begin{gathered} w=(1)/(2)(D-U) \\ =(1)/(2)(20-12) \\ =(1)/(2)*8 \\ =4\text{ miles/h} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/pm7b0wmgtulusaeg0p6fqy3ybygvidnmbr.png)
Therefore, the speed of the boat in still water is 16 mph .
The speed of the current is 4 miles mph.