a motorboat takes 4 hours to travel 128 kilometers going up stream. the return trip takes 2 hours going downstream. what is the rate of the boat in still water and what is the rate of the current?

Answer :

Answer:

The rate of the boat in still water is;

[tex]48\text{ km/h}[/tex]

The rate of the current is;

[tex]16\text{ km/h}[/tex]

Explanation:

Let x represent the rate of the boat in still water, and y represent the rate of the current;

Speed upstream is;

[tex]x-y[/tex]

Speed downstream is;

[tex]x+y[/tex]

Recall that distance equals the product of speed and time;

Upstream the product is;

[tex]4(x-y)=128------1[/tex]

Downstream;

[tex]2(x+y)=128-------2[/tex]

Expand both equation and multiply equation 2 through by 2;

[tex]\begin{gathered} 4x-4y=128--------1a \\ 2x+2y=128 \\ \text{ multiply through by 2;} \\ 4x+4y=256------2a \end{gathered}[/tex]

solve the simultaneous equation by elimination; add equation 1a and 2a together;

[tex]\begin{gathered} 4x+4x+4y-4y=128+256 \\ 8x=384 \\ x=\frac{384}{8} \\ x=48 \end{gathered}[/tex]

Then we can substitute x into equation equation 2 to get y;

[tex]\begin{gathered} 2(x+y)=128 \\ x+y=\frac{128}{2} \\ y=64-x \\ y=64-48 \\ y=16 \end{gathered}[/tex]

Therefore, the rate of the boat in still water is;

[tex]48\text{ km/h}[/tex]

The rate of the current is;

[tex]16\text{ km/h}[/tex]

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