- #1
-Dragoon-
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First of all, I'd like to thank all the kind and wonderful people here who have helped me so far in concepts I've been struggling with. I greatly appreciate it. I'd also like to wish every a happy and wonderful new year!
Find the displacement of each of the situations by adding the vectors:
a) 12.5 km [E], 7.4 km [W]
b) 8.0 cm [N], 6.0 cm [W]
c) 2.0 m [E 20° S], 4.0 m
d) 100 km, 80km [ W 30° S], 20 km [N].
C^2 = a^2 + b^2
C^2 = a^2 + b^2 - 2abCosC
SinA/a = SinB/b = SinC/c
Alright so today I started learning the very basics of adding vector quantities. I find parts a and b relatively easy. I know that solving part a is just changing the 7.4 km [W] vector into -7.4 km [E]. For part b, I would draw the triangle and apply pythagorean theorem and finding the direction would be opp/adj = tanΘ.
Now it is part c where I start to really struggle. I've only been introduced to one method of solving for both situations in part c and d where specific directions have been given, and I don't believe it to be the most effective. First I would find displacement by applying the equation: c^2 = a^2 + b^2 - 2(a)(b)Cos(C), and I know that C would be the difference of the direction and 90°. Using the unit circle, since it is going clock-wise in the southern direction of east, that is -20°. So C would then be 70 degrees, correct? After finding the displacement, I would find the direction by applying the sine equation:SinA/a = SinB/b = SinC/c. But, for the last part I have no conventional method I can use that I believe would be effective. Any help? Is there an easier way or more effective method of solving part C, if possible? I don't have a teacher, as I do this course through correspondence, so it is quite difficult making sense of one or two examples provided in my course book.
Thanks for all the help, everyone and have a happy new year!
Homework Statement
Find the displacement of each of the situations by adding the vectors:
a) 12.5 km [E], 7.4 km [W]
b) 8.0 cm [N], 6.0 cm [W]
c) 2.0 m [E 20° S], 4.0 m
d) 100 km
Homework Equations
C^2 = a^2 + b^2
C^2 = a^2 + b^2 - 2abCosC
SinA/a = SinB/b = SinC/c
The Attempt at a Solution
Alright so today I started learning the very basics of adding vector quantities. I find parts a and b relatively easy. I know that solving part a is just changing the 7.4 km [W] vector into -7.4 km [E]. For part b, I would draw the triangle and apply pythagorean theorem and finding the direction would be opp/adj = tanΘ.
Now it is part c where I start to really struggle. I've only been introduced to one method of solving for both situations in part c and d where specific directions have been given, and I don't believe it to be the most effective. First I would find displacement by applying the equation: c^2 = a^2 + b^2 - 2(a)(b)Cos(C), and I know that C would be the difference of the direction and 90°. Using the unit circle, since it is going clock-wise in the southern direction of east, that is -20°. So C would then be 70 degrees, correct? After finding the displacement, I would find the direction by applying the sine equation:SinA/a = SinB/b = SinC/c. But, for the last part I have no conventional method I can use that I believe would be effective. Any help? Is there an easier way or more effective method of solving part C, if possible? I don't have a teacher, as I do this course through correspondence, so it is quite difficult making sense of one or two examples provided in my course book.
Thanks for all the help, everyone and have a happy new year!