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Published byFrancis Curtis Modified over 8 years ago
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In the last section, we dealt with two vectors in the same direction, opposite directions, and at right angles to each other. In this section we will deal with one vector, and try to figure out the other two sides of the triangle.
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1. draw a coordinate system (a graph) 2. choose the correct formula 3. plug in your numbers (calculator must be in degree mode!) 4. check to see if your answer needs a negative sign (what quadrant is your vector in?)
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For the X side use Ax = Acos θ For the Y side use Ay = Asin θ
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Draw a vector of a car that travels 60 km to the northwest
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Draw a vector representing a football being thrown 30 m south of east at a 60° angle (that’s physics speak for southeast)
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60°
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When you are given a direction to draw your vector, simply follow the directions; north of east, south of east, north of west or south of west.
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If you are only given an angle, start measuring your angle from the positive x axis and move counterclockwise.
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The angle you use in a calculation is always the angle INSIDE the triangle you formed
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Draw a picture showing a golf ball being hit at an angle of 110° Answer?...
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What are the components of a vector of magnitude 5.6 m at an angle of 47° from the positive x axis?
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47°
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Because “solving for the components” means what are the x and y sides of my triangle, I will use Ax = Acos θ for the x side, and Ay = Asin θ for the y side Here’s how this looks… Ax = 5.6cos47 ( Ax = 3.82m) Ay = 5.6sin47 (Ay = 4.09m)
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Since the original vector is in quadrant 1, no.
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P. 74 #’s 11-14 P. 79 #’s 25, 26, 27a, 29 (use graph paper or be very careful!) and 30
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