To determine the magnitude of the resultant velocity of the boat, given its components in the x and y directions, we can use the Pythagorean theorem. Here's a detailed, step-by-step solution:
1. Identify the velocities in the x and y directions:
- The boat's velocity in the y-direction ([tex]\(v_y\)[/tex]) is [tex]\(15.0 \ \text{m/s}\)[/tex].
- The current's velocity in the x-direction ([tex]\(v_x\)[/tex]) is [tex]\(4.00 \ \text{m/s}\)[/tex].
2. Understand that these velocities form a right triangle:
- One leg of the triangle is the velocity in the y-direction [tex]\(v_y = 15.0 \ \text{m/s}\)[/tex].
- The other leg of the triangle is the velocity in the x-direction [tex]\(v_x = 4.00 \ \text{m/s}\)[/tex].
3. Apply the Pythagorean theorem to find the resultant velocity ([tex]\(v\)[/tex]):
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the resultant velocity, [tex]\(v\)[/tex]) is equal to the sum of the squares of the lengths of the other two sides (the velocities [tex]\(v_x\)[/tex] and [tex]\(v_y\)[/tex]).
Therefore:
[tex]\[
v^2 = v_x^2 + v_y^2
\][/tex]
4. Substitute the given values:
[tex]\[
v^2 = (4.00 \ \text{m/s})^2 + (15.0 \ \text{m/s})^2
\][/tex]
5. Perform the calculations:
[tex]\[
v^2 = 4.00^2 + 15.0^2
\][/tex]
[tex]\[
v^2 = 16 + 225
\][/tex]
[tex]\[
v^2 = 241
\][/tex]
6. Take the square root of both sides to solve for [tex]\(v\)[/tex]:
[tex]\[
v = \sqrt{241}
\][/tex]
7. Find the numerical value:
[tex]\[
v \approx 15.524 \ \text{m/s}
\][/tex]
Thus, the magnitude of the boat's velocity is approximately [tex]\( 15.524 \ \text{m/s} \)[/tex].
