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igomit [66]
3 years ago
5

A boat makes a 120-mile trip downstream in 3 hours but makes the return trip in 4 hours. if b = the rate of the boat in still wa

ter and c = the rate of the current, which of the following represents the trip up stream
Mathematics
1 answer:
Lyrx [107]3 years ago
3 0

Answer:

Step-by-step explanation:

The boat makes 120 mile trip downstream in 3 hours, but makes the return trip in 4 hours. If b-the rate of the boat in still water and c= the rate of the current, the following equations represents the trip upstream; 4(b - c) = 120. The answer will be B.

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Answer:

x =\dfrac{45 \sqrt{6}}{ 2}

Step-by-step explanation:

From the given information:

The diagrammatic interpretation of what the question is all about can be seen in the diagram attached below.

Now, let V(x) be the time needed for the runner to reach the buoy;

∴ We can say that,

\mathtt{V(x) = \dfrac{70-x}{7}+\dfrac{\sqrt{54^2+x^2}}{5}}

In order to estimate the point along the shore, x meters from B, the runner should  stop running and start swimming if he want to reach the buoy in the least time possible, then we need to differentiate the function of V(x) and relate it to zero.

i.e

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-\dfrac{1}{7}+ \dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}=0

\dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}

\dfrac{5x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}

\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{\dfrac{7}{5}}

\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{5}{7}

squaring both sides; we get

\dfrac{x^2}{54^2+x^2}= \dfrac{5^2}{7^2}

\dfrac{x^2}{54^2+x^2}= \dfrac{25}{49}

By cross multiplying; we get

49x^2 = 25(54^2+x^2)

49x^2 = 25 \times 54^2+ 25x^2

49x^2-25x^2 = 25 \times 54^2

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x^2 = \dfrac{25 \times 54^2}{24}

x =\sqrt{ \dfrac{25 \times 54^2}{24}}

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x =\dfrac{270}{\sqrt{4 \times 6}}

x =\dfrac{45 \times 6}{ 2 \sqrt{ 6}}

x =\dfrac{45 \sqrt{6}}{ 2}

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