Answer:
speed of the boat = 10.67 m/s...
direction the boat is 38.2 degrees north of east...
Step-by-step explanation:
To solve this problem, we can use vector addition.
Let the velocity of the boat be represented by vector B, which has a magnitude of 12 m/s and is directed 45 degrees north of east. We can represent this vector as:
B = 12 m/s at 45 degrees
Let the velocity of the ocean current be represented by vector C, which has a magnitude of 2 m/s and is directed directly south. We can represent this vector as:
C = 2 m/s at 270 degrees
To find the actual speed of the boat and the direction it is heading, we need to add these two vectors together using vector addition. To do this, we can break each vector into its x and y components:
Bx = 12 m/s * cos(45) = 8.49 m/s
By = 12 m/s * sin(45) = 8.49 m/s
Cx = 0 m/s
Cy = -2 m/s
Now, we can add the x and y components of the vectors separately:
Rx = Bx + Cx = 8.49 m/s + 0 m/s = 8.49 m/s
Ry = By + Cy = 8.49 m/s - 2 m/s = 6.49 m/s
The resulting vector R has a magnitude of:
|R| = sqrt(Rx^2 + Ry^2) = sqrt((8.49 m/s)^2 + (6.49 m/s)^2) = 10.67 m/s
And a direction of:
theta = arctan(Ry/Rx) = arctan(6.49 m/s/8.49 m/s) = 38.2 degrees
Therefore, the actual speed of the boat is 10.67 m/s and the direction the boat is heading is 38.2 degrees north of east...