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Due Sun 05/26/2019 11:59 pm Show Intro/Instnactios At noon, ship A is 20 nautical miles due west of ship B. Ship A...


Due Sun 05/26/2019 11:59 pm Show Intro/Instnactios At noon, ship A is 20 nautical miles due west of ship B. Ship A is sailing
Due Sun 05/26/2019 11:59 pm Show Intro/Instnactios At noon, ship A is 20 nautical miles due west of ship B. Ship A is sailing west at 25 knots and ship B is sailing nerth at 23 knets. How fas ( knoes) is the distance betwen the ships changing at6 PMT (Nete 1 knot is a speed of I nautical mile per hout) knots Preview knt Get help Video Points possble: This is attempt 1 of Subit
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After t hours

The position of Ship A will be (20 + 25*t, 0)

The position of Ship B will be (0, 23*t)

Hence, the distance between the two ships at any time t will be

s = \sqrt{(20 + 25t - 0)^2 + (0 - 23t)^2}

Taking the derivative of the s function with respect to t, we get

ds 2(20 25t) (25) + 2(23t) (23)

1154t 500 aS dtV/((20 + 25t)2 + (23t)2)

Substituting the value of t=6, since time difference between 6 pm and 12 noon is equal to 6

33.90 knots dt

Hence the distance is changing wrt time at the rate of 33.90 knots when time is equivalent to 6 hours

Note - Post any doubts/queries in comments section.

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Due Sun 05/26/2019 11:59 pm Show Intro/Instnactios At noon, ship A is 20 nautical miles due west of ship B. Ship A...
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