Question #155191

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?


1
Expert's answer
2021-01-14T17:53:38-0500

The distance between ships will be

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

here t=0 at 2pm.


s(t)=210t10+2(5+5t)52(10t)2+(5+5t)2s'(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


210t10+2(5+5t)5=02*10t*10+2*(5+5t)*5=0

200t+50+50t=0200t+50+50t=0

250t=50250t=-50

t=15t=-\frac{1}{5}


To find moment of time:

2pm15=1.48pm2pm-\frac{1}{5}=1.48pm


Answer: 1.48pm.


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