Question

One ship is sailing south at a rate of 5 knots, and another is sailing east at a rate of 10 knots. At 2 P.M. the second ship was at the place occupied by the first ship one hour before. At what time does the distance between the ships not changing?

Answer 1

So at 2:00 pm the first ship would be 5 nautical miles directly south of the second ship.

Let the time passes since 2:00 pm be t hours

So the two ships form a right-angled triangle, with a horizontal side of 10t nmiles and a vertical of (5+5t) nmiles
Let the distance between them, the hypotenuse , be D nmiles
D^2 = (10t)^2 + (5+5t)^2
when the distance D is not changing, dD/dt = 0
2D dD/dt = 2(10t)(10) + 2(5+5t)(5)
200t + 50 + 50t = 0
250t = -50
t = -50/250 = -1/5 or -12 minutes

so the distance did not change at 1:48 pm

Answer 2

Why did it become 5+5t?

Answer 3

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