Answer:
8.07 m/s, 81.7º NE.
Explanation:
- The ship, due to the local ocean current, will be deviated from its original due north bearing.
- In order to find the magnitude of the velocity of the ship, we need to convert a vector equation, in an algebraic one.
- If we choose two axes coincident with the N-S and W-E directions, we can find the components of the velocity along these directions.
- Clearly, the velocity of the ship, relative to water, is only due north, so it has no component along the W-E axis.
- The local ocean current, as it is directed at an angle between both axes, has components along these axes.
- These components can be found from the projections of the velocity vector along these axes, as follows:
[tex]vocx = voc* cos 40 = 1.53 m/s * 0.766 = 1.17 m/s\\vocy = voc* sin 40 = 1.53 m/s * 0.643 = 0.98 m/s[/tex]
- The component along the N-S axis (y-axis) of the velocity of the ship will be the sum of the velocity relative to water, plus the component of the ocean current along this same axis:
[tex]vshy = vsw + vwy = 7.00 m/s + 0.98 m/s = 7.98 m/s[/tex]
- The component along the W-E axis, is just the component of the local ocean current in this direction:
vshx = 1.17 m/s
- We can find the magnitude of the velocity vector, applying the Pythagorean theorem, as follows:
[tex]v = \sqrt{vshx^{2} + vshy^{2} } =\sqrt{(7.98m/s)^{2} +(1.17m/s)^{2} } =8.07 m/s[/tex]
- The direction of the vector relative to the W-E axis (measured in counterclockwise direction) is given by the relative magnitude of the x and y components, as follows:
[tex]tg \theta = \frac{vshy}{vshx} = \frac{7.98}{1.17} = 6.82 \\ \theta = tg^{-1} (6.82)\\ \theta= 81.7\deg[/tex]
- The velocity of the ship, relative to Earth, is 8.07 m/s, 81.7º North of East.
