A river flows due East at 0.979 m/s. A boat crosses the river from the South shore to the North shore by maintaining a constant ve-

locity of 12.8 m/s due North relative to the
water. The river is 306 m wide.
What is the velocity of the boat relative to
shore? Answer in units of m/s.

I visualize a right triangle:

velocity:sqrt(12.8^2+.979^2)

now for velocity, you need an angle;

theta off of the N direction= arctan .979/12.8

idk you tell me...

To find the velocity of the boat relative to the shore, we need to combine the velocity of the boat in still water with the velocity of the river.

The velocity of the boat in still water is given as 12.8 m/s due North relative to the water. This means that if there were no river flow, the boat would be moving at a constant velocity of 12.8 m/s directly North.

The velocity of the river is given as 0.979 m/s due East.

To combine these velocities, we need to use vector addition. The boat's velocity relative to the shore will be the vector sum of its velocity in still water and the velocity of the river.

Since the boat is moving North and the river is flowing East, these velocities form a right triangle. The width of the river is the base of the triangle, and the velocity of the boat in still water is the vertical side. The resultant velocity (velocity of the boat relative to the shore) will be the hypotenuse of the triangle.

We can use the Pythagorean theorem to solve for the magnitude of the resultant velocity:

(Resultant velocity)^2 = (velocity of the boat in still water)^2 + (velocity of the river)^2

Let's substitute the given values into the equation:

(Resultant velocity)^2 = (12.8 m/s)^2 + (0.979 m/s)^2

(Resultant velocity)^2 = 163.84 m^2/s^2 + 0.958441 m^2/s^2

(Resultant velocity)^2 = 164.798441 m^2/s^2

Taking the square root of both sides gives us:

Resultant velocity = √(164.798441 m^2/s^2)

Resultant velocity ≈ 12.85 m/s

Therefore, the velocity of the boat relative to the shore is approximately 12.85 m/s.