An ocean liner positioned at longitude 15^\circ\text{W} records its local solar time as 1:… — JAMB UTME

An ocean liner positioned at longitude $15^\circ\text{W}$ records its local solar time as 1:00 p.m. At the exact same moment, a cargo ship located at a different meridian records its local solar time as 5:00 p.m. What is the longitudinal position of the cargo ship?

$45^\circ\text{E}$ Answer
B
$75^\circ\text{W}$
C
$60^\circ\text{E}$
D
$45^\circ\text{W}$

Answer

The longitude of the cargo ship is

45^\circ\text{E}

Because the cargo ship's local time is 4 hours ahead of the ocean liner's time (5:00 p.m. vs 1:00 p.m.), it must be located east of the liner. Since Earth rotates

15^\circ

per hour, a 4-hour difference corresponds to an angular distance of

60^\circ

. Measuring

60^\circ

east from

15^\circ\text{W}

involves traveling

15^\circ

east to the Prime Meridian (

0^\circ

) and then another

45^\circ

east into the Eastern Hemisphere, placing the cargo ship at

45^\circ\text{E}

Step-by-Step Solution

Calculate the time difference between the two vessels.

Time difference =

\text{5:00 p.m.} - \text{1:00 p.m.} = 4\text{ hours}

Determining the time interval is necessary to find the total angular distance.

Convert the time difference into degrees of longitude.

Angular distance =

4\text{ hours} \times 15^\circ/\text{hour} = 60^\circ

Earth rotates

360^\circ

in 24 hours, which equals

15^\circ

per hour.

Determine the direction of movement and calculate the target longitude.

Since 5:00 p.m. is later than 1:00 p.m., the cargo ship is to the East. Starting at

15^\circ\text{W}

, moving

15^\circ

East reaches

0^\circ

(Greenwich Meridian), and moving the remaining

45^\circ

(

60^\circ - 15^\circ

) East reaches

45^\circ\text{E}

Locations with later local times are located further East.

Key Concept

Longitude and Local Time Calculations across Meridians

Estimated Time:1m 30s

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