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Sagot :
It is given that the distance is 6 miles and the time is 2 hours upstream and one and a quarter hour downstream.
The time downstream is given by:
[tex]1\frac{1}{4}=\frac{4+1}{4}=\frac{5}{4}\text{ hours}[/tex]Since the distance is constant, it follows:
[tex]\begin{gathered} \text{ Speed=}\frac{\text{ Distance}}{\text{ Time}} \\ \text{ Distance=SpeedxTime} \end{gathered}[/tex]So the distance is constant hence:
[tex]\text{ Distance upstream=Distance Downstream}[/tex]Let the speed of kayak be x and speed of current be y so the speed downstream is x+y and speed upstream is x-y so it follows:
[tex]\begin{gathered} \frac{5}{4}(x+y)=2(x-y) \\ 4\times\frac{5}{4}(x+y)=4\times2(x-y) \\ 5x+5y=8(x-y) \\ 5x+5y=8x-8y \\ 5y+8y=8x-5x \\ 13y=3x \\ x=\frac{13}{3}y \end{gathered}[/tex]Use the individual equation to find x and y as follows:
[tex]\begin{gathered} 6=2(x-y) \\ 6=2(\frac{13}{3}y-y) \\ 3=\frac{13-3}{3}y \\ \frac{9}{10}=y \end{gathered}[/tex]Hence the speed of the water current is 9/10 miles per hour.
The speed of the kayak is given by:
[tex]\begin{gathered} x=\frac{13}{3}y \\ x=\frac{13}{3}\times\frac{9}{10} \\ x=\frac{39}{10}=3.9\text{ miles per hour} \end{gathered}[/tex]Hence the speed of the kayak without the water current is 3.9 miles per hour.
The time required without water current is:
[tex]\begin{gathered} \text{Time}=\frac{Dis\tan ce}{Speed} \\ t=\frac{6}{3.9}\approx1.5\text{ hours} \end{gathered}[/tex]Hence it will take approximately 1.5 hours without the current.
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