SOLUTION: a) In this question, i is a unit vector due east and j is a unit vector due north. A cyclist rides at a speed of 4ms−1 on a bearing of 015°. Write the velocity vector of the

Algebra ->  Test -> SOLUTION: a) In this question, i is a unit vector due east and j is a unit vector due north. A cyclist rides at a speed of 4ms−1 on a bearing of 015°. Write the velocity vector of the       Log On


   



Question 1200933: a) In this question, i is a unit vector due east and j is a unit vector due north.
A cyclist rides at a speed of 4ms−1 on a bearing of 015°. Write the velocity vector of the cyclist in
the form xi + yj, where x and y are constants. [2]
(b) A vector of magnitude 6 on a bearing of 300° is added to a vector of magnitude 2 on a bearing of
230° to give a vector v. Find the magnitude and bearing of v.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
a) In this question, i is a unit vector due east and j is a unit vector due north.
A cyclist rides at a speed of 4ms−1 on a bearing of 015°. Write the velocity vector of the cyclist in the form xi + yj, where x and y are constants.
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x = 4*sin(15)
y = 4*cos(15)
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(b) A vector of magnitude 6 on a bearing of 300° is added to a vector of magnitude 2 on a bearing of 230° to give a vector v. Find the magnitude and bearing of v.
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Use the Cosine Law to find the magnitude of r, the resultant:
r^2 = 6^2 + 2^2 - 1*6*2*cos(110)
r =~ 6.6411
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Use the Law of Sines to find the angle at the Origin:
6.6411/sin(110) = 2/sin(A)
sin(A) = 2*sin(110)/6.6411 = ~0.28299A = ~16.439 degs
Bearing = A + 300 = 316.439 degs