Given:
Number of hours, t = 14 hours
Distance = 52 miles
Speed travelling = 8 mph
Let's find the speed of the current.
Apply the formula:
![\begin{gathered} \text{ speed=}(distnace)/(time) \\ \\ time=(distance)/(speed) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/8clvv86h92nao0hx1v5zkpluyrtv18gl53.png)
Here, we have the system of equations:
For time travelling up: 14 = 52/8-c
For time travelling down: 14 = 52/8 + c
Where c is the speed of the current.
Hence, we have:
![(52)/(8-c)+(52)/(8+c)=14](https://img.qammunity.org/2023/formulas/mathematics/high-school/c7bz712yhj26gq1o8czc29fq0hmpjs3mid.png)
Let's solve the equation for c.
Multiply all terms by (8-c)(8+c):
![\begin{gathered} (52)/(8-c)(8-c)(8+c)+(52)/(8+c)(8-c)(8+c)=14(8-c)(c+c) \\ \\ 52(8+c)+52(8-c)=14(8-c)(8+c) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/e8wvemb9a7b3786uybzxti6dqsyi5brptx.png)
Solving further, expand using FOIL method and apply distributive property:
![\begin{gathered} 52(8)+52c+52(8)-52c=14(64-c^2) \\ \\ 416+52c+416-52c=896-14c^2 \\ \\ 416+416+52c-52c=896-14c^2 \\ \\ 832-896=-14c^2 \\ \\ -64=-14c^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/2xr2k5bv1c4kw1kybeuos9ezio1jsjk70r.png)
Solving further:
Divide both sides by -14
![\begin{gathered} (-64)/(-14)=(-14c^2)/(-14) \\ \\ 4.57=c^2 \\ \\ c^2=4.57 \\ \\ \text{ Take the square root of both sides:} \\ √(c^2)=√(4.57) \\ \\ c=2.14 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ox4gx7qd32ehbhqxei3eo70h52b92ztwwn.png)
Therefore, the speed of the current was 2.14 miles per hour.
ANSWER:
2.14 mph