A ferry boat departed from port and traveled at a rate of ten miles per hour at an angle of 30 degrees north of east across a large river. If the current is flowing at a rate of 4 miles per hour straight south, what is the speed and direction of the boat in the water? Round any irrational answers to one decimal place.

the east component is ... 10 cos(30º)

the north component is
... 10 sin(30º) - 4

speed = √[(east)^2 + (north)^2]

tan(Θ) = north / east

Θ is degrees north of east

To find the speed and direction of the boat in the water, we need to use vector addition.

Let's break down the given information into vectors.

The speed of the boat relative to the water is 10 miles per hour. Let's call this vector B, with magnitude 10 mph.

The current is flowing at a rate of 4 miles per hour straight south. We can represent this as vector C, with magnitude 4 mph and direction directly downwards.

Now, let's draw a sketch to visualize this scenario:

B (10 mph, 30° North of East)





▼ C (4 mph, pointing South)

To find the resulting vector D, which represents the speed and direction of the boat in the water, we can use the cosine law for vector addition:

|D|^2 = |B|^2 + |C|^2 - 2|B||C|cos(θ)

where |D| is the magnitude of vector D, |B| is the magnitude of vector B, |C| is the magnitude of vector C, and θ is the angle between vectors B and C.

First, let's calculate the values:

|B| = 10 mph
|C| = 4 mph
θ = 30°

Plugging these values into the equation, we have:

|D|^2 = (10 mph)^2 + (4 mph)^2 - 2(10 mph)(4 mph)cos(30°)

Simplifying:

|D|^2 = 100 + 16 - 80cos(30°)

Now, let's calculate cos(30°):

cos(30°) ≈ 0.866

Substituting the value into the equation, we have:

|D|^2 = 100 + 16 - 80(0.866)
|D|^2 ≈ 100 + 16 - 69.28
|D|^2 ≈ 46.72

Taking the square root of both sides, we find:

|D| ≈ √46.72
|D| ≈ 6.8 mph (rounded to one decimal place)

Therefore, the speed of the boat in the water is approximately 6.8 mph.

To determine the direction, we can use trigonometry.

In the given scenario, the boat is traveling at an angle of 30 degrees north of east.

To find the direction of the boat in the water, we need to consider the angle it makes with the positive x-axis.

Since 30 degrees is north of east, the angle relative to the positive x-axis is 90° - 30° = 60°.

Therefore, the direction of the boat in the water is 60 degrees east of the positive x-axis.

Hence, the speed of the boat in the water is approximately 6.8 mph, and it is traveling at an angle of 60 degrees east of the positive x-axis.