Question
Asked Dec 16, 2019
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At noon, a ship leaves a harbor and sails south at 10 knots. Two hours later, a second ship leaves the harbor and sails east at 15 knots. When will the ships be 100 nautical miles apart? Round to the nearest minute.

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Expert Answer

Step 1

Given:

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Let v, = 10 knots the speed of the first ship. Let v, = 15 knots the speed of the second ship. And t be the time of the first ship, t – 2 be the time of second ship. And d =100 nautical miles. It is known that Distance = speed x time %3D Thus, d, = 10r and d, =15(t – 2)

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Step 2

By using Pythagore...

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(107)* + (15(t – 2))* = (100) 100r? +(15t – 30) = 10,000 100t? + 225t? +900 – 900t = 10,000 325t? – 900t – 9,100 = 0 13t? – 36t – 364 = 0 -(36) + (-36) – 4(13)(-364) 2.13 2(9±4/79) 13 2(9– 4/79 2(9+4/79 or r t = 13 13 t = 6.85 or t=-4.085(Rejected)

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