Two ships leave port at 4 p.m. One
is headed N38°E and is travelling at
11.5 km/h. The other is travelling at
13 km/h, with heading S47°E. How far
apart are the two ships at 6 p.m.?
time for each is 2 hours
Did you make a sketch?
Mine is a triangle with sides 23 and 26 with a contained angle of 95°.
Looks like clear case of using the cosine law.
To determine how far apart the two ships are at 6 p.m., we need to calculate the distance traveled by each ship during this time period.
Ship 1 (N38°E) is traveling at 11.5 km/h. Since 4 p.m. to 6 p.m. is a 2-hour time period, the distance traveled by Ship 1 can be calculated using the formula:
Distance = Speed × Time
Distance (Ship 1) = 11.5 km/h × 2 hours = 23 km
Ship 2 (S47°E) is traveling at 13 km/h. Using the same formula:
Distance (Ship 2) = 13 km/h × 2 hours = 26 km
Now, we can calculate the distance between the two ships using the Pythagorean theorem in a right triangle:
Distance² = (Distance (Ship 1))² + (Distance (Ship 2))²
Distance² = 23² + 26²
Distance² = 529 + 676
Distance² = 1205
Taking the square root of both sides gives us:
Distance = √1205 ≈ 34.7 km
Therefore, the two ships are approximately 34.7 km apart at 6 p.m.
To find the distance between the two ships at 6 p.m., we need to determine the positions of both ships at that time.
Let's start by calculating the distance traveled by each ship from 4 p.m. to 6 p.m.
Ship 1:
Speed = 11.5 km/h
Time = 6 p.m. - 4 p.m. = 2 hours
Distance = Speed * Time = 11.5 km/h * 2 hours = 23 km
Ship 1 has traveled 23 km.
Ship 2:
Speed = 13 km/h
Time = 6 p.m. - 4 p.m. = 2 hours
Distance = Speed * Time = 13 km/h * 2 hours = 26 km
Ship 2 has traveled 26 km.
Now, we can use these distances and the given headings to find the positions of the two ships.
For Ship 1:
Heading = N38°E
This means Ship 1 is moving northeast, with an angle of 38° from the north direction.
To find the horizontal and vertical components of the distance traveled by Ship 1, we can use trigonometry.
Horizontal distance = Distance * cos(angle)
Vertical distance = Distance * sin(angle)
Horizontal distance = 23 km * cos(38°) ≈ 18.19 km
Vertical distance = 23 km * sin(38°) ≈ 14.09 km
For Ship 2:
Heading = S47°E
This means Ship 2 is moving southeast, with an angle of 47° from the south direction.
To find the horizontal and vertical components of the distance traveled by Ship 2, again we use trigonometry.
Horizontal distance = Distance * cos(angle)
Vertical distance = -Distance * sin(angle) [negative because Ship 2 is moving south]
Horizontal distance = 26 km * cos(47°) ≈ 17.33 km
Vertical distance = -26 km * sin(47°) ≈ -18.89 km
Now, let's find the total horizontal and vertical distances between the two ships by adding up the corresponding components.
Total horizontal distance = Horizontal distance of Ship 2 - Horizontal distance of Ship 1
= 17.33 km - 18.19 km ≈ -0.86 km
Total vertical distance = Vertical distance of Ship 2 - Vertical distance of Ship 1
= -18.89 km - 14.09 km ≈ -32.98 km
The negative sign indicates that the two ships are moving away from each other in opposite directions.
Finally, we can use the Pythagorean theorem to find the distance between the two ships:
Distance = √(Total horizontal distance² + Total vertical distance²)
= √((-0.86 km)² + (-32.98 km)²)
≈ √(0.7396 km² + 1087.0404 km²)
≈ √(1087.780 km²)
≈ 32.96 km
Hence, the two ships are approximately 32.96 km apart at 6 p.m.