Who used less sugar 5/10 or 5/6
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Based on the information provided, = at a bearing angle of = 59.88° .
We are requested to calculate the overall displacement of the ship, both in magnitude and direction, once it departs the harbor under the stated parameters. First, let's define a bearing.
<h3>What is bearing in the context given above?</h3>Angles are generally measured anticlockwise from the positive x-axis, whereas bearing angles are evaluated clockwise from the positive y-axis. A bearing is NOT a standard angle measuring tool.
Therefore, a bearing of 44.0° indicates that this is an angle 90.0° - 44.0° = 46°. This is the angle to be used in the calculations.
We're given that the first displacement is 11.4° at an angle of 46° (which was computed earlier).
Step 1 - Split the above into components
x1 = 11.4 Cos 46°
= 7.92m
y1 = 11.4Sin 46.0°
= 11.4 * 0.71933980033
= 8.2m
The second displacement is a simple 6.2mi due east, that is, the positive
x-direction. The components are thus:
Δx = x1 + x2
= 7.92 + 6.2
= 14.12mi
Δy = 1 + y2
= 8.2 + 0
= 8.2
r = √(x total)² + (y total)²
=√[(14.21)²+(8.2)²]
= √(201.9241 +67.24)
= √269.1641
= 16.4mi
The direction of the displacement vector is given by:
tan Ф = (Δy)/(Δx)
= arctan (8.2/14.12)
= arctan (0.5807365439)
= 30.12°
Recall that we were asked for the bearing angle. The bearing angle is what we get when we subtract 30.12° from 90°.
That is:
= 90 - 30.12
= 59.88°
Learn more about bearing angle:
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