What is the angle north of east in this kinematics problem?

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The discussion revolves around a kinematics problem involving a walk of 6.0 km in a direction north of east, resulting in a final position of 2.0 km east and an unknown distance north. The correct angle north of east is determined to be 71 degrees. Participants emphasize the importance of drawing a diagram to visualize the problem, which helps in applying trigonometric functions to find the angle. The solution involves using the cosine function to relate the sides of the triangle formed by the walk. Overall, the discussion highlights the utility of visual aids and trigonometry in solving kinematics problems.
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Physics Kinematic Problem -- need help

Well here I am studying for the final and this question pops up and I have no idea how to do it. I know the answer however, but it does not really help me.

If you walk 6.0 km in a straight line in a direction north of east and you end up 2.0 km east and several kilometers north. How many degrees north of east have you walked?

Choices:

a) 19 degrees
b) 45 degrees
c) 60 degrees
d) 71 degrees

Answer: d

Thank you for your help
 
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The trig to this is to try drawing the walk on a piece of paper.
 
Did you draw a diagram? I drew a triangle with a vertex at the origin and the hypotenuse going north and east (up and to the right) for 6 units. The opposite side would be 2 units...Can you finish from there?
 
would it be possible to draw it in the forum?? I still don't get it. I knew how to do this a while ago and it was pretty easy. I did draw a diagram but it didn't really help for some reason. Maybe I read the question wrong, but either way I still don't understand. :cry:

AH HA! Thanks Fred. When you said 'opposite' I thought it meant opposite side of the plane. Ah it means the opposite side. Ok thanks I get it.

Cos -1 (2/6) = 71

:D

Thanks again
 
Last edited:
This is simple. First you walk 6km north of east, you end up 2km east and some km's north . Draw a triangle representing the above values where base=2 and hypotenuse=6

therefore,

cosQ= \frac{1}{3}
 
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