At noon, ship A starts sailing due east at the rate of 20km/hr. Ath the time, ship B, which is located 100km east of ship A initially, starts sailing on a course 60 degrees north of west at the rate of 10km/hr. How fast is the distance between the two ships changing one hour after they start sailing?

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To find the rate at which the distance between the two ships is changing, we can use the concept of relative velocity.

Let's consider the position of ship A after one hour. It would have traveled 20 km due east.

The position of ship B after one hour can be calculated by using the velocity vector for ship B. The magnitude of the velocity vector is 10 km/hr, and the angle is 60 degrees north of west. Since north is perpendicular to east, this means the angle formed by the velocity vector is (90 - 60) = 30 degrees west of north.

Using trigonometry, we can decompose the velocity vector into its components:
Vertical component = 10 km/hr * sin(30 degrees)
Horizontal component = 10 km/hr * cos(30 degrees)

The distance between the two ships after one hour can be calculated using the Pythagorean theorem:
Distance = √((100 km + 20 km)² + (10 km/hr * cos(30 degrees))² + (10 km/hr * sin(30 degrees))²)

To find the rate at which the distance is changing, we can take the derivative of the distance with respect to time (t):
d(Distance)/dt = (1/2) * (2 * (100 km + 20 km) * (0 km/hr) + 2 * (10 km/hr * cos(30 degrees) * (-sin(30 degrees) * (1/2)) + 2 * (10 km/hr * sin(30 degrees) * (cos(30 degrees) * (1/2))

Simplifying, we get:
d(Distance)/dt = (10 km/hr * cos(30 degrees) * (-sin(30 degrees)) + 10 km/hr * sin(30 degrees) * (cos(30 degrees))

Evaluating the expression using trigonometric values:
d(Distance)/dt = (10 km/hr * (sqrt(3)/2) * (-1/2) + 10 km/hr * (1/2) * (sqrt(3)/2))
= 10 km/hr * (-sqrt(3)/4 - sqrt(3)/4)
= 10 km/hr * (-sqrt(3)/2)

Therefore, the distance between the two ships is changing at a rate of -10√3 km/hr (negative meaning the distance is decreasing) one hour after they start sailing.

To find the rate at which the distance between the two ships is changing, we can use the concept of relative velocity. Here's how to approach the problem step by step:

1. Draw a diagram to visualize the situation. Draw a line representing the east-west direction, with ship A starting at the origin (0 km) and ship B located 100 km east of ship A initially (at the point x = 100 km).

2. Set up a coordinate system. Let the x-axis represent the east-west direction, with the positive x-axis pointing east, and the y-axis represent the north-south direction, with the positive y-axis pointing north.

3. Define the variables. Let the position of ship A be (x₁, y₁) and the position of ship B be (x₂, y₂). The distance between the two ships is given by the formula:

d = sqrt((x₂ - x₁)² + (y₂ - y₁)²)

4. Determine the velocities of ship A and ship B. Ship A is sailing due east at a constant speed of 20 km/hr, so its velocity vector can be written as:

v₁ = (20 km/hr, 0 km/hr)

Ship B is sailing 60 degrees north of west at a speed of 10 km/hr. To convert this velocity vector into x and y components, use trigonometry. The x-component of the velocity is given by:

v₂x = -10 km/hr * cos(60°)

The negative sign indicates that the velocity is directed to the west. The y-component of the velocity is given by:

v₂y = 10 km/hr * sin(60°)

5. Determine the positions of the ships after one hour. Ship A would have traveled a distance of 20 km in the east direction, so its new position is (20 km, 0 km). Ship B would have traveled a distance of 10 km in the direction given by the vector (v₂x, v₂y), so its new position is (x₂ + v₂x, y₂ + v₂y).

6. Use the distance formula to find the distance between the ships after one hour. Substitute the positions of the ships into the distance formula:

d = sqrt((x₂ + v₂x - x₁)² + (y₂ + v₂y - y₁)²)

7. Calculate the rate at which the distance is changing. To find how fast the distance between the two ships is changing after one hour, take the derivative of the distance equation with respect to time:

dd/dt = (d/dt) [sqrt((x₂ + v₂x - x₁)² + (y₂ + v₂y - y₁)²)]

8. Substitute the values into the equation. Plug in the values for v₂x, v₂y, x₁, x₂, y₁, and y₂ obtained from the given information in the problem.

9. Simplify and calculate the final answer.

By following these steps, you will be able to calculate the rate at which the distance between the two ships is changing one hour after they start sailing.