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?


Expert's answer

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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