To find the resultant velocity of the boat, we treat the boat’s velocity and the river’s velocity as vectors. The boat is moving due south, and the river is moving due east, so we have a right-angled triangle situation.

Given:

• The velocity of the river, $v_r = 1.5 \, \text{m/s}$ (east).
• The velocity of the boat, $v_b = 2.0 \, \text{m/s}$ (south).

We can find the resultant velocity using the Pythagorean theorem since these velocities are perpendicular:

$v_{\text{resultant}} = \sqrt{v_r^2 + v_b^2}$

Substitute the given values:

$v_{\text{resultant}} = \sqrt{(1.5)^2 + (2.0)^2}$ $v_{\text{resultant}} = \sqrt{2.25 + 4.00}$ $v_{\text{resultant}} = \sqrt{6.25}$ $v_{\text{resultant}} = 2.5 \, \text{m/s}$

Thus, the magnitude of the resultant velocity is $2.5 \, \text{m/s}$.

Direction of the Resultant Velocity

To find the direction (angle $\theta$) relative to due south, we use the tangent function:

$\tan(\theta) = \frac{v_r}{v_b} = \frac{1.5}{2.0} = 0.75$

Now, take the inverse tangent:

$\theta = \tan^{-1}(0.75) \approx 36.87^\circ$

So, the resultant velocity makes an angle of approximately $36.87^\circ$ east of south.

Final Result

• Magnitude of the resultant velocity: $2.5 \, \text{m/s}$
• Direction: $36.87^\circ$ east of south.

Would you like more details on any part of the solution?

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Here are 5 related questions:

1. What happens to the resultant velocity if the boat's speed increases to 3.0 m/s?
2. How would the resultant velocity change if the river flow slows to 1.0 m/s?
3. Can we calculate the time taken by the boat to cross the river?
4. What is the total distance traveled by the boat while crossing the river?
5. How would the result differ if the boat aimed straight across the river instead of heading due south?

Tip: Drawing a vector diagram helps visualize and solve problems involving motion in two dimensions.