A rowboat is heading across a river at 2.4 m/s east. The current of the river is 1.7 m/s south. What is the rowboat's resultant velocity?

Draw the two vectors. The resultant speed is the hypotenuse of the triangle.

That is, √(2.4^2 + 1.7^2) = 2.94 m/s

The direction is arctan(1.7/2.4) = E35.3°S

To find the rowboat's resultant velocity, we need to use vector addition.

First, let's consider the rowboat's velocity heading east. This velocity has a magnitude of 2.4 m/s in the east direction (positive x-axis).

Next, let's consider the current of the river. The current has a velocity of 1.7 m/s in the south direction (negative y-axis).

To add these two velocities together, we need to consider their directions. Since the rowboat's velocity is in the east direction and the current is in the south direction, we can think of them as forming a right triangle. The rowboat's velocity is the horizontal leg, and the current's velocity is the vertical leg.

Using the Pythagorean theorem, we can find the magnitude of the resultant velocity:

Resultant velocity = √((2.4 m/s)^2 + (-1.7 m/s)^2)

Calculating the squares and summing them:

Resultant velocity = √(5.76 m^2/s^2 + 2.89 m^2/s^2)
Resultant velocity = √(8.65 m^2/s^2)
Resultant velocity ≈ 2.94 m/s

So, the rowboat's resultant velocity is approximately 2.94 m/s. To specify its direction, we need to find the angle it makes with the positive x-axis.

Using trigonometry, we can find this angle:

θ = arctan((-1.7 m/s) / (2.4 m/s))

Calculating the arctan value:

θ ≈ -37.97°

The negative sign indicates that the angle is measured clockwise from the positive x-axis.

Therefore, the rowboat's resultant velocity is approximately 2.94 m/s in a direction approximately 37.97° clockwise from the positive x-axis.