A ship is headed due north at a constant 20 miles per hour. Because of the ocean current, the true course of the ship is 15°. If the currents are a constant 18 miles per hour, in what direction are the currents running? (Enter your answers as a comma-separated list. Round your answers to the nearest whole number.)
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To determine the direction of the currents, we can use vector addition.
Let's consider the ship's velocity as one vector, v_ship, with a magnitude of 20 miles per hour and a direction due north.
The true course of the ship is 15°, which means it is deviating 15° to the east of the north direction. We can represent this deviation as another vector, v_course, with magnitude 20 miles per hour and direction 15° east of north.
The combined effect of the ship's velocity and the currents will be the resultant vector, v_resultant. The magnitude of the resultant vector can be found using the Law of Cosines:
|v_resultant|^2 = |v_ship|^2 + |v_current|^2 - 2 * |v_ship| * |v_current| * cos θ
where θ is the angle between v_ship and v_current. In this case, θ is the angle between the true course of the ship (v_course) and the currents (v_current).
Let's substitute the given values into the equation:
|v_resultant|^2 = 20^2 + 18^2 - 2 * 20 * 18 * cos 15°
Simplifying the equation gives:
|v_resultant|^2 = 400 + 324 - 720 * cos 15°
Calculating the right side of the equation gives:
|v_resultant|^2 ≈ 400 + 324 - 720 * 0.9659
≈ 400 + 324 - 697.608
≈ 26.392
Taking the square root of both sides, we find:
|v_resultant| ≈ √26.392
≈ 5.137 miles per hour
The magnitude of the resultant vector is approximately 5.137 miles per hour.
Now, let's find the direction of the currents. Since the ship is deviating 15° to the east of north, and the currents take the ship further in that direction, we can subtract 15° from the direction of the resultant vector.
So, the direction of the currents is approximately (15° - 15°) = 0°.
Therefore, the currents are running due north.
To find the direction of the currents, we need to examine the vector components of the ship's velocity and the ocean current.
Let's break down the ship's velocity into its northward (N) and eastward (E) components. Since the ship is headed due north, its N component will be 20 miles per hour, and its E component will be 0 miles per hour.
Now, let's break down the ocean current velocity into its N and E components. Since the true course of the ship is 15°, the current is pushing the ship slightly to the east. We can use trigonometry to determine the components of the current velocity.
The current velocity can be broken down into two components: one parallel to the ship's direction (E component) and one perpendicular to the ship's direction (N component). The E component can be found by multiplying the current velocity (18 mph) by the cosine of the angle between the true course and due north (15°). The N component can be found by multiplying the current velocity by the sine of the angle between the true course and due north.
E component = 18 mph * cosine(15°)
N component = 18 mph * sine(15°)
Calculating these values, we find:
E component ≈ 17.18 mph
N component ≈ 4.64 mph
Therefore, the direction of the currents is approximately eastward (E), and the magnitude is about 17 mph. Putting it all together, the direction of the currents can be represented as follows: 90° (north) + 90° (east) ≈ 180°. So, the direction of the currents can be rounded to 180°.
Hence, the direction of the currents is 180 degrees.