Thread:RePeat/@comment-1693968-20140305220535/@comment-1693968-20140309034324

Danielboone6702 wrote: Okay, I'm going to do a detailed explanation of the problem :

''The brigantine was sailing at full speed. He was 30 miles at sea when Lord Nelson began to give chase at twice the speed of the brigantine. If Lord Nelson caught up in six hours, what was the brigantine's speed?''

First off, we need to find out when the two people catch up, i.e., when they are the same distance away from the finish line. The formula for distance in terms of speed and time is this :

v*t + c = d, where v is velocity (or speed), t is time, c is head start, and d is distance.

For example, if you went 60 mph for 2 hours, then you would go 120 miles, plus any head start you may have had. Pretty straightforward to understand.

So, we need to come up with distance formulas for both people and find out when they equal each other. We know both people's head starts (0 for Nelson, and 30 miles for the brigantine), and the time that they will travel (6 hours), but we don't know their speeds. However, we do know that Nelson goes twice as fast as the brigantine, so let's call the brigantine's speed x, in which case Nelson's speed would be 2x. Let's write the formulas :

(Brigantine's distance) x * 6 + 30 = 2x*6 (Nelson's distance)

Simplify: 6x + 30 = 12x

Subtract 6x from both sides: 30 = 6x

Divide both sides by 6: x = 5

And you have the answer: The brigantine's speed was 5 mph! Drew, you mentioned that you could have solved this equation in an easier way, but that was because one person went twice a fast as another. But what if one person went 3 times as fast as the other, or even pi times as fast as the other? That method would not work. This method is a far more general way to solve these types of problems. Yeah, I'm aware it only works because of the situation. However, if you'll notice, I was just using logic to quickly simplify the problem as much as possible. Logic is easily applicable to all problems, and is especially useful for ones of this difficulty. If you're getting into Physics, and even Algebra 2, that kind of logic can become much more inefficient. In that case, you would have to use the equations you're taught. But for this specific problem, it's much quicker to just solve it the way I did.