Two ships leave a port at 9 a.m. One travels at a bearing of N 53° W at 12 miles per hour, and the other travels at a bearing of S 67° W at 19 miles per hour. Approximate how far apart they are at noon that day. (Round your answer to one decimal place.)
To approximate how far apart the two ships are at noon, we can use the concept of relative velocity. The relative velocity is the vector sum of the velocities of the two ships.
First, let's determine the time duration between 9 a.m. and noon. Since there are 3 hours between 9 a.m. and noon, the time duration is 3 hours.
Now, we need to calculate the displacement of each ship during this time duration.
For the ship traveling at a bearing of N 53° W at 12 miles per hour, the displacement can be calculated using the formula:
displacement = velocity * time
displacement = 12 miles/hour * 3 hours = 36 miles
For the ship traveling at a bearing of S 67° W at 19 miles per hour, the displacement can also be calculated using the same formula:
displacement = velocity * time
displacement = 19 miles/hour * 3 hours = 57 miles
Next, we can use these displacements to find the distance between the two ships. We can treat the displacements as the sides of a triangle and use the Law of Cosines to find the distance.
The Law of Cosines states that, in a triangle with sides a, b, and c, and angle C opposite side c:
c^2 = a^2 + b^2 - 2ab * cos(C)
In our case, let's assume the 36-mile displacement is side a, the 57-mile displacement is side b, and the angle between these two displacements is 180° (since they are in opposite directions).
c^2 = 36^2 + 57^2 - 2 * 36 * 57 * cos(180°)
Simplifying:
c^2 = 1296 + 3249 - 2 * 36 * 57 * (-1)
c^2 = 1296 + 3249 + 4104
c^2 = 8649
Taking the square root of both sides:
c ≈ √8649
c ≈ 93.037 miles
Therefore, the two ships are approximately 93.0 miles apart at noon that day.