A ship leaves port at 2:00 pm and sails in the direction S34 degrees W at a rate of 15mph. Another ship leaves port at 3:00 pm and sails in the direction S56 degrees E at a rate of 12mph. Find how far apart the ships are at 4:00 pm.
no idea whatcha got?
Convert each length traveled to an x & y displacement.
Then use the law of cosines to find the third length.
To find the distance between the ships at a particular time, we can use the concept of relative velocity.
Step 1: Convert the given angles to compass bearings.
- S34 degrees W can be converted to S56 degrees E by subtracting 90 degrees from the given angle.
- S56 degrees E remains the same.
Step 2: Convert the compass bearings to Cartesian coordinates.
- S34 degrees W can be represented as (-sin(34 degrees), -cos(34 degrees)).
- S56 degrees E can be represented as (sin(56 degrees), -cos(56 degrees)).
Step 3: Calculate the displacement of each ship using the given rates and the time difference between their departures.
- The first ship departs at 2:00 pm and the second ship departs at 3:00 pm, so the time difference is 1 hour.
- The displacement of the first ship is (15 mph * 1 hour) * (-sin(34 degrees), -cos(34 degrees)).
- The displacement of the second ship is (12 mph * 1 hour) * (sin(56 degrees), -cos(56 degrees)).
Step 4: Find the distance between the ships by calculating the magnitude of the vector difference between their displacements.
- The distance between the ships is the magnitude of ((15 * -sin(34) - (12 * sin(56))) , (-15 * cos(34) + 12 * -cos(56))).
Step 5: Simplify the expression and calculate the magnitude of the resulting vector.
- The distance between the ships at 4:00 pm is the magnitude of ((-10.01), (-19.46)).
- Using the distance formula |AB| = √((x2 - x1)^2 + (y2 - y1)^2), we get |AB| = √((-10.01)^2 + (-19.46)^2).
Step 6: Calculate the final answer.
- |AB| ≈ √(100.20 + 378.85).
- |AB| ≈ √479.05.
- |AB| ≈ 21.89 miles (rounded to two decimal places).
Therefore, the ships are approximately 21.89 miles apart at 4:00 pm.