Solving vectors

[b]1. A person going for a walk follows a path. The total trip consists of four straight-line paths. At the end of the walk what is the person’s resultant displacement measured from the starting point? The books says the answer is 240 m at 237 degrees.

[b] Vector A is 100m East at 0°, Vector B is 300m south and vector c is 150 meters southwest and vector D is 200 meters north west. There are two angle measurements given. I dont know how to upload the exact picture but if you go to http://ift.tt/1glYC3c and find the exact problem, the picture is on there as well as the answer. I dont know how they got to that answer. Please help.

[b]3.A_x=100m
A_y=0m
B_x=0m
B_y=300m
C_x=150cos(120)=-75m
C_y=150sin(120)=129.9 m
d_x=200cos(75)=51.8 m?
d_y=200sin(75)=193.2

Im not sure how to get the correct angle measurements for vector d. I tried to figure it out from the angles given but it was very confusing.

to find the resultant I did F_x=100+0+0+-75=25m
F_y=193.2+129.9+300+0=569.1m
[itex]/sqrt{F_x^2+F_y^2}
i got 569.6m
and tan^-1(569.1/25)=87.5°

Where did I go wrong? I think it is the angle measurements I put in but Im not sure how to get the correct measurements.

http://ift.tt/1iZqCtJ

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