A ship is travelling due east at 10 m/s. What must be the speed of a second ship heading 30 degrees east of north if it is always due north of the first ship?
To determine the speed of the second ship, we need to break down the velocity vectors of both ships into their respective components.
Let's consider the first ship traveling due east at 10 m/s. Since the ship is traveling due east, its velocity vector has no northward component, only an eastward component. Therefore, the velocity vector of the first ship can be represented as (10 m/s, 0 m/s) in component form.
Now, let's calculate the velocity vector of the second ship. The second ship is always due north of the first ship but at a 30-degree angle east of north. This means the northward component of the velocity vector will be equal to the eastward component multiplied by the tangent of the angle.
Given that the eastward component of the first ship's velocity vector is 10 m/s, the northward component of the second ship's velocity vector will be:
Northward component = Eastward component × tan(angle)
= 10 m/s × tan(30°)
Using a scientific calculator, we find the value of tan(30°) is approximately 0.5774. Therefore:
Northward component = 10 m/s × 0.5774
≈ 5.774 m/s
Since the second ship is only traveling in the northward direction, its velocity vector can be represented as (0 m/s, 5.774 m/s) in component form.
To find the speed of the second ship, we can use the Pythagorean theorem. The speed is the magnitude of the velocity vector, which can be calculated as follows:
Speed of the second ship = √((Northward component)^2 + (Eastward component)^2)
= √((5.774 m/s)^2 + (0 m/s)^2)
= √(33.311 m^2/s^2)
≈ 5.77 m/s
Therefore, the speed of the second ship heading 30 degrees east of north, always due north of the first ship traveling due east at 10 m/s, is approximately 5.77 m/s.