Answer to Question #155191 in Calculus for Phyroe

Question

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

Answer 1

The distance between ships will be

$s(t)=\sqrt{(10t)^2+(5+5t)^2}$

here t=0 at 2pm.

$s'(t)=\frac{2*10t*10+2(5+5t)*5}{2\sqrt{(10t)^2+(5+5t)^2}}$

the distance was minimum when s'(t)=0, and in this moment

the distance between ships was not changing, so

$2*10t*10+2*(5+5t)*5=0$

$200t+50+50t=0$

$250t=-50$

$t=-\frac{1}{5}$

To find moment of time:

$2pm-\frac{1}{5}=1.48pm$

Answer: 1.48pm.