A boat left dock A, traveled north for 10 miles, then 45 East of north for 20 miles and docked at B.

A.How far north did the boat travel? How far east did it travel? B. Find the magnitude and direction of the direct-path vector AB.
a. North 24.1 East 14.1
b. 22.4 magnitude 45 East of North

b can NOT be 45 degrees and the magnitude MUST be bigger than 24.1

north = 10 + 20 cos 45 = 24.1 agree
east = 20 sin 45 = 14.1 agree

mag = sqrt (24.1^2+14.1^2) = 27.94
tan deg e of n = 14.1/24.1
so deg e of n = 30.3 degrees East of North

To find the distances traveled by the boat and the magnitude and direction of the direct-path vector AB, we can use basic trigonometry and vector operations.

A. To determine how far north the boat traveled, we need to find the vertical component of the boat's displacement. The boat initially traveled north for 10 miles, so the vertical component is 10 miles.

To determine how far east the boat traveled, we need to find the horizontal component of the boat's displacement. The boat traveled 45 degrees East of North for 20 miles. To find the horizontal component, we need to use trigonometric functions.

The horizontal component can be calculated using the formula: horizontal component = displacement * cos(angle)

In this case, the displacement is 20 miles and the angle is 45 degrees East of North.

Horizontal component = 20 miles * cos(45 degrees)
Horizontal component = 20 miles * cos(45 degrees)
Horizontal component ≈ 20 miles * 0.7071
Horizontal component ≈ 14.1 miles

Therefore, the boat traveled approximately 10 miles north and 14.1 miles east.

B. To find the magnitude and direction of the direct-path vector AB, we can use the Pythagorean theorem and inverse trigonometric functions.

The magnitude of a vector can be found using the formula: magnitude = √(horizontal component^2 + vertical component^2)

In this case, the horizontal component is 14.1 miles and the vertical component is 10 miles.

Magnitude = √(14.1 miles^2 + 10 miles^2)
Magnitude = √(198.81 miles^2 + 100 miles^2)
Magnitude ≈ √298.81
Magnitude ≈ 17.3 miles

The direction of the vector can be found using the inverse tangent function. The angle can be calculated using the formula: angle = arctan(vertical component / horizontal component)

In this case, the vertical component is 10 miles and the horizontal component is 14.1 miles.

Angle = arctan(10 miles / 14.1 miles)
Angle ≈ arctan(0.7092)
Angle ≈ 34.5 degrees

Therefore, the magnitude of the direct-path vector AB is approximately 17.3 miles and its direction is 34.5 degrees East of North.