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The problem is:

Given two boats, one, the coast-guard boat, stands on the point zero, the thief's boat, on the point $D$, the coast-guard boat travels in a speed of $Vg$ knots and the thief's boat, in $Vf$ knots, Given $D$, $Vf$ and $Vg$, say if it's possible to the coast guard boat reach the thiefs boat before the thief's boat reach $12$ miles away from the point $0$

I've started with an Inequalitie $0 + (i*Vg) <= D + (i*Vf)$ with $D + (i*Vf) < 12 miles$ being $i$ a moment. but know I'm stuck on how to continue, and how to get a formula to solve it.

Thanks

EDIT:

Sorry for the inconvenience, I've just misunderstood the statement, they aren't perpenticular to the cost when they leave, imagine as a cartesian plane, the coast-guard boat is in the point $(0,0)$ and the thief, in the point $(d, 0)$ and the objective of the thief is reach the point $(d, 12)$ before the coast guard boat reach him, it gives an all new view to to problem, that's why my example in the comments isn't wrong..

Sorry for the incovenience.

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    @LeonidKovalev sorry, it was an error due my bad english, thanks2012-07-13

1 Answers 1

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Presumably the thief will go directly away from the origin, so it takes $\frac {12-D}{Vf}$ hours to reach the point $12$ miles away. In that time the Coast Guard boat can go $\frac {(12-D)Vg}{Vf}$ miles. If this is at least $12$, it will catch the thief. So your condition is $\frac {(12-D)Vg}{Vf} \ge 12$. The question did not seem to ask when the Coast Guard catches the thief, but if you want the time $i$, you are close in that it comes when $iVg=D+iVf$, which gives $i=\frac D{Vg-Vf}$

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    @aajjbb: If $Vf=Vg$, the coast guard never catches the thief. You have the inequality backwards. It would call for $\frac {(12-10)5}5 \ge 12$ or $2 \ge 12$ which is not true, so the thief is not caught.2012-07-13