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A submarine left the Dania Pier and traveled toward a navigational buoy at an average speed of 16 mph. A cruise ship left three hours later and traveled in the same direction but with an average speed of 28 mph. Find the number of hours the submarine traveled before the cruise ship caught up.

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Final answer:

The submarine traveled for a total of 9 hours before the cruise ship caught up to it, considering the head start and relative speeds of both vehicles.

Step-by-step explanation:

The question involves determining the number of hours a submarine traveled before being caught up to by a cruise ship, given their respective speeds and the head start time of the submarine. To solve this, we use the concept of relative speed and time. Since the submarine has a head start, we first calculate the distance it covers in those three hours, which is 48 miles (16 mph × 3 hours). When the cruise ship starts, it closes the gap at a relative speed of 28 mph - 16 mph = 12 mph.

Let the time taken for the cruise ship to catch up to the submarine be t hours. Then, the distance covered by the cruise ship in that time is 28 mph × t, and the distance covered by the submarine in the same time is 16 mph × t. To close a gap of 48 miles, the distances traveled by both must satisfy the equation: (16 × t) + 48 = 28 × t. Solving for t gives us t = 6 hours. Therefore, the submarine travels for 3 + 6 = 9 hours in total before being caught up to by the cruise ship.

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