A boat is heading due east at a constant speed of 35 mi/h. There is an 8 mi/h current moving north. What is the boat’s actual speed and direction?

Let x=east, y=north,

x-component of velocity = 35
y-component of velocity = 8
Magnitude=sqrt(8^2+35^2)=sqrt(1289)=35.9 mph. approx.
direction:
θ=tan-1(35/8)
direction=NθE

thanks so much

You're welcome!

To determine the boat's actual speed and direction, we need to use vector addition.

First, we need to convert the given speed and current into vectors. The boat is heading due east, so its velocity vector is 35 mi/h in the east direction. The current is moving north, so its velocity vector is 8 mi/h in the north direction.

Next, we add the two velocity vectors to find the boat's actual velocity vector. Since the vectors are at right angles to each other, we can use the Pythagorean theorem to find the magnitude of the resulting vector, and trigonometry to find its direction.

The magnitude of the resulting vector (R) can be calculated using the formula R = sqrt((V1^2) + (V2^2)), where V1 and V2 are the magnitudes of the individual velocity vectors.

substituting the given values, R = sqrt((35^2) + (8^2)) = sqrt(1225 + 64) = sqrt(1289) ≈ 35.93 mi/h.

To find the direction, we use trigonometry. Since the east direction and north direction are at right angles, we have a right triangle. The angle (θ) between the boat's velocity vector and the east direction is given by the equation tan(θ) = V2 / V1, where V1 is the magnitude of the east velocity vector and V2 is the magnitude of the north velocity vector.

Substituting the given values, tan(θ) = 8 / 35, using the inverse tangent function (arctan), we can find θ ≈ 13.11 degrees.

Therefore, the boat's actual speed is approximately 35.93 mi/h and its direction is approximately 13.11 degrees north of east.