Your vessel departs Seattle at 1010 zone...

Question 1 of 2707066636eea28f7522a1c51b7aa

Your vessel departs Seattle at 1010 zone time, (ZD +8), on 28 May bound for Apra, Guam, (ZD -10). The distance by great circle is 4,948 miles, and you estimate that you will average 18.5 knots. What is your estimated zone time of arrival?

Back

Question 1 of 27070

🔍 Key Concepts

• Speed–time–distance relationship for sea passage: $Time = \frac{Distance}{Speed}$ • How to convert travel time in hours into days, hours, and minutes • How to correctly apply zone description (ZD) changes when moving from ZD +8 to ZD -10 and cross the date line area

💭 Think About

• First, ignore time zones and just compute how long the trip will take in hours using the distance and average speed. What total time in days, hours, and minutes do you get? • Add that elapsed time to the departure date and time in Seattle zone time. What tentative arrival (date/time) do you get before any ZD correction? • Now compare the departure ZD (+8) to the arrival ZD (-10). In going from +8 to -10, are you effectively moving the clock forward or backward, and by how many hours total? How does that affect the date?

✅ Before You Answer

• Be sure you convert knots to hours correctly: 1 knot = 1 nautical mile per hour • Double-check the difference in ZD between +8 and -10. Is it 18 hours or 16 hours, and in which direction do you apply it? • After you adjust for the ZD change, verify that your final date matches one of the multiple-choice options, and that the time portion ends with :37 as indicated by all options.