at noon a ship leaves a harbor and sails south at 50 knots t

at noon, a ship leaves a harbor and sails south at 50 knots. three. hours layer, a second ship leaves the harbor and sails east at 20 knots. when will the ships be 400 nautical miles apart? round to nearest minute
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nnect Sadion 1.5 Wppicabores al Quadratic Question 12 at 12 point) At noon, athip leaves a barber and ails wuth leaves the harbor and sail 10 kaott. When the thips be Eautical apart Reund to the nearwd The ships will be nautical miles apart a

Solution

Let the two ships be 400 nautical miles apart after x hours after noon. In x hours, the 1^st ship would travel 50 x nautical miles and in (x-3) hours, the 2^nd ship would travel 20(x-3) hours. Since East and South directions are perpendicular to each other, and since the two ships are 400 nautical miles apart, hence by Pythagores theorem, (400)² = (50x)² + [20(x-3)]² or, 160000 = 2500x² +400(x² -6x+9) = 2500x² +400x² -2400x +3600 or, 2900x² -2400x +3600- 160000 = 0 or, 2900x² -2400x – 156400 = 0 or,on dividing both the sides by 100, we get 29x² -24x-1564 = 0. Now, on using the qudratic formula, we get x = [-(-24)±[(-24)²-4*29*(-1564)}]/2*29= [24 ±(576+181424)]/58 = [24±182000]/58 = (24± 426.6146)/58. Now, since time cannot be negative, we have x = (24+ 426.6146)/58 = 450.6146/28 = 7.76922 hours or, 7 hours, 46 minutes (approximately). Thus the two ships will be 400 nautical miles apart , 7 hours, 46 minutes after the 1^st ship leaves the harbour.