Vector addition and displacement

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MIA6
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Homework Statement


During a rodeo, a clown runs 8.0 m north, turns 55 degrees north of east, and runs 3.5 m. Then, after waiting for the bull to come near, the clown turns due east and runs 5.0 m to exit the arena. What is the clown's total displacement?

The Attempt at a Solution


I drew a pic according to the problem, and it looks like a trapezoid, i don't know if it is the right image. How can i find the total displacement's two components? By adding three vectors together? Please explain it to me, thank you.
 
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MIA6 said:
but is this a trapezoid?

It might have two sides that are parallel, hard to tell but I don't think that would help you even if it was...
 
Try resolving each vector into its horizontal and vertical components, and then add them.
 
ok, what's 8.0m's x and y component? its y-component is itslef? the same problem with 5m. hope you can give me some details since i really have no clue.
 
MIA6 said:
ok, what's 8.0m's x and y component? its y-component is itslef? the same problem with 5m. hope you can give me some details since i really have no clue.

8.0 meters north means that this displacement has no x component and a y component of + 8.0 meters.

Break into x and y components all three displacement vectors and add all the x and all the y components.
 
Well the 8m is directly north, so expressed in terms of its components, it would be <0,8>, so yes, its y component is itself, and its x component is zero. The 5m vector will be <5,0>, since it has no y component.
 
You could also turn it into a right triangle and solve with the law of sines only. There are quite a few ways to solve for the total displacement.
 
MIA6 said:
ok, what's 8.0m's x and y component? its y-component is itslef? the same problem with 5m. hope you can give me some details since i really have no clue.


Assuming that North is the positive y-axis and East is the positive x-axis:
Since 8.0m points due north, its x-component is zero. All it is is a y-component.