To find the resultant velocity of the boat, we can consider the velocities as vectors and use the Pythagorean theorem because the boat's path is perpendicular to the river's current.
The boat is heading due east with a velocity of 10 m/s, and the river flows due north at 6 m/s. We can break this down as follows:
- Eastward velocity (V_east) = 10 m/s
- Northward velocity (V_north) = 6 m/s
Now, we can find the resultant velocity (V_resultant) using the Pythagorean theorem:
\[ V_{\text{resultant}} = \sqrt{(V_{\text{east}})^2 + (V_{\text{north}})^2} \] \[ V_{\text{resultant}} = \sqrt{(10, \text{m/s})^2 + (6, \text{m/s})^2} \] \[ V_{\text{resultant}} = \sqrt{100 + 36} \] \[ V_{\text{resultant}} = \sqrt{136} \] \[ V_{\text{resultant}} \approx 11.62, \text{m/s} \]
Next, to find the direction of the resultant velocity, we can calculate the angle (θ) relative to the eastward direction using the tangent function:
\[ \tan(\theta) = \frac{V_{\text{north}}}{V_{\text{east}}} \] \[ \tan(\theta) = \frac{6}{10} = 0.6 \]
To find the angle θ, we take the arctangent:
\[ \theta = \tan^{-1}(0.6) \approx 30.96^\circ \]
Thus, the resultant velocity of the boat is approximately 11.62 m/s at an angle of 30.96° north of east.