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ETA. Question #5 on page 104

Denny Orewiler Nov 06, 2017 01:43PM EST

I am doing quite well on figuring the ETA questions beginning on page 103 in Navigation for the Mariner.
However, number 5 escapes me. I get a distance of 26.2 miles traveled at a speed of 12.6 NM. Time traveled comes out to 2 hours and 5 minutes. Added to 1815, I get an ETA of 2020. The correct answer is 2032. Am I using the wrong buoy (Plum Island Mid Channel buoy)?

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MLS Staff Instructor Nov 06, 2017 04:24PM EST Mariners Learning System Agent

All of the solutions can be found in your online classroom:

Plotting Solution:

You will find the distance between the two fixes (points A and B) is 28.7 nautical miles in this problem.

The problem gives the speed of the vessel as 12.6 knots.

Using the information we now know set up a DST table. Fill in the distance to be traveled (28.7 nm).

Fill in the speed of the vessel as 12.6 knots.

Take the distance to be traveled (28.7 nm) and divide it by the speed of the vessel (12.6 knots). The resulting number is 2.277 hours. Since we only need two places after the decimal point we will round this number to 2.28 hours.This number does not represent 2 hours and 28 minutes it represents 2.28 hours.

Next you must convert this decimal to minutes. To make the conversion simply multiply your fraction of hours (.28 hours) by 60 minutes. You will get 16.8 minutes. (16.8 minutes represents 16 minutes and 48 seconds). We are only working with whole minutes so we will round 16.8 minutes to 17 minutes. We will find that 2.28 hours is the same as 2 hours and 17 minutes.

As a last step in the solution add the 2 hours and 17 minutes to your position A time (1815) and you will find that your ETA to a position 5 miles due south of Falkner Island Light to be 2032.

Answer C is the correct response.

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