Question

It is 6 miles in a kayak to the Fish Islands from my house. The trip to the island takes 2 hourstraveling against the current and 1¼ hours for the return trip (with the current). How fast can Ipaddle the Kayak if there was no current? The answer can be rounded to the nearest tenth.Solve Algebraically using linear systems

Answer 1

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:

`1(1)/(4)=(4+1)/(4)=(5)/(4)\text{ hours}`

Since the distance is constant, it follows:

`\begin{gathered} \text{ Speed=}\frac{\text{ Distance}}{\text{ Time}} \\ \text{ Distance=SpeedxTime} \end{gathered}`

So the distance is constant hence:

`\text{ Distance upstream=Distance Downstream}`

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:

`\begin{gathered} (5)/(4)(x+y)=2(x-y) \\ 4*(5)/(4)(x+y)=4*2(x-y) \\ 5x+5y=8(x-y) \\ 5x+5y=8x-8y \\ 5y+8y=8x-5x \\ 13y=3x \\ x=(13)/(3)y \end{gathered}`

Use the individual equation to find x and y as follows:

`\begin{gathered} 6=2(x-y) \\ 6=2((13)/(3)y-y) \\ 3=(13-3)/(3)y \\ (9)/(10)=y \end{gathered}`

Hence the speed of the water current is 9/10 miles per hour.

The speed of the kayak is given by:

`\begin{gathered} x=(13)/(3)y \\ x=(13)/(3)*(9)/(10) \\ x=(39)/(10)=3.9\text{ miles per hour} \end{gathered}`

Hence the speed of the kayak without the water current is 3.9 miles per hour.

The time required without water current is:

`\begin{gathered} \text{Time}=(Dis\tan ce)/(Speed) \\ t=(6)/(3.9)\approx1.5\text{ hours} \end{gathered}`

Hence it will take approximately 1.5 hours without the current.