1 Answer

Ibrahim Azhar Armar · Answer 1 · 2021-10-29T09:48:12+0000

Answer:

8.07 m/s, 81.7º NE.

Step-by-step 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:

$vocx = voc* cos 40 = 1.53 m/s * 0.766 = 1.17 m/s\\vocy = voc* sin 40 = 1.53 m/s * 0.643 = 0.98 m/s$

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:

vshy = vsw + vwy = 7.00 m/s + 0.98 m/s = 7.98 m/s

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:

$v = \sqrt{vshx^(2) + vshy^(2) } =\sqrt{(7.98m/s)^(2) +(1.17m/s)^(2) } =8.07 m/s$

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:

$tg \theta = (vshy)/(vshx) = (7.98)/(1.17) = 6.82 \\ \theta = tg^(-1) (6.82)\\ \theta= 81.7\deg$

The velocity of the ship, relative to Earth, is 8.07 m/s, 81.7º North of East.

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