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# word problemm

two fishing boats leave the harbor at the same time and using the same float plan. one average 17 knots (nautical miles per hour), and the other average 14.5 knots. how long will it take for the boats to be 10 nautical miles apart?

Because the boats left at the same time, you can use the time they've been traveling since they've left the harbor as any variable (I'll use 't' here).

The first boat is traveling at a rate of 17 knots, so the equation for distance of the first boat from the harbor after t hours will be 17*t. Simply put, this means that you can plug in how long the boat has been traveling into t to see how far away it is from the harbor.

The second boat is traveling at a rate of 14.5 knots, so again the equation of the distance from the harbor will be 14.5*t after t hours.

Because you want to know how long it will take for the boats to be 10 nautical miles apart, you can model that as the difference of the two rates. Think of it as Distance of boat 1 - Distance of boat 2 = 10 nautical miles.
Remember that each distance is in nautical miles because the distance equation uses (nautical miles/hour)*(hours), where the hours cancel out, leaving you with nautical miles as your resulting unit.

If you continue to solve this problem, you get:

17*t - 14.5*t = 10

Solve for t:
2.5*t = 10
t = 10/2.5
t = 4.

Therefore after 4 hours, the boats will be 10 nautical miles apart.

2/24/2013

Let t = the time it takes the boats to be 10 nautical miles apart. Since rate x time = distance, the first boat's distance would be 17x and the second boat's distance would be 14x.

Since the first boat would be 10 nautical miles further, then we can express the relationship as

14x + 10 = 17x

Combining like terms, we get

10 = 3x

x = 10/3 hours or  3 1/3 hours

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