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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.



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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Statements of the form $a\le b$ and $a<b$ are inequalities. The term inequation is sometimes used for $a\ne b$. – user31373 Jul 13 '12 at 0:54
@LeonidKovalev sorry, it was an error due my bad english, thanks – aajjbb Jul 13 '12 at 1:01
up vote 1 down vote accepted

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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Yes, think a little more I've got this formula, but other doubt, Think of the test case with $D = 9$, $Vf = 12$ and $Vg = 15$, working this entry, the result is $3.75$ and due to my book, it considers that the coast-guard boat reaches the thief, but it don't fulfill the inequality. – aajjbb Jul 13 '12 at 12:17
@aajjbb: The Coast Guard can only go $3.75$ miles in the $\frac 14$ hour before the thief reaches mile $12$, so fails. That is in accord with the inequality. The final equation says it will be 3 hours until the Coast Guard catches the thief, at which point they will be 45 miles away. It is all consistent, so I would say the book is wrong the thief gets caught in this case. – Ross Millikan Jul 13 '12 at 13:12
yes, I've misunderstood the example, in this case they are caught really, but there are more tricky cases, it give me more examples, like $D = 10$, $Vf = 5$ and $Vg = 5$ in this case, the inequality is fulfill with a result of $2 <= 12$, but in the book, in this case, the thief is not caught, weird not ? – aajjbb Jul 13 '12 at 13:43
@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. – Ross Millikan Jul 13 '12 at 13:49

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