To find the magnitude of the boat's resultant velocity vector and its direction, you can use vector addition. The boat's velocity consists of two components: one due north at 10 mph and another 65° west of north at 5 mph. We'll break this down into its components:
1. Northward component (N):
N = 10 mph
2. Westward component (W):
W = 5 mph * cos(65°)
Now, let's calculate the westward component (W):
W = 5 mph * cos(65°)
W ≈ 5 mph * 0.4226183
W ≈ 2.1131 mph
Now, you have the northward and westward components of velocity. To find the resultant vector (R), you can use the Pythagorean theorem:
R^2 = N^2 + W^2
R^2 = (10 mph)^2 + (2.1131 mph)^2
R^2 = 100 + 4.4691
R^2 = 104.4691
Now, take the square root:
R ≈ √104.4691
R ≈ 10.22 mph (rounded to the nearest hundredth)
So, the magnitude of the boat's resultant velocity vector is approximately 10.22 mph.
To find the direction, you can use the inverse tangent (arctan) function:
Direction (θ) = arctan(W / N)
Direction (θ) = arctan(2.1131 mph / 10 mph)
Direction (θ) ≈ arctan(0.21131)
Direction (θ) ≈ 11.61 degrees (rounded to two decimal places)
Therefore, the boat's resultant vector has a magnitude of approximately 10.22 mph, and it's directed at an angle of approximately 11.61 degrees west of north.