A dog searching for a bone walks 3.50 m south, then 8.20 m at an angle of 30.0 degrees north of east, and finally 15.0 m west. Use graphical techniques to find the dog's resultant displacement vector.
The answer in the back of the book says 7.9 m at 4.3 north of west...
please explain every step thoroughly!
2 Answers

Okay, this will take some work on your part so that you best understand it.
First draw both axes. Once that is done, draw a vector of (theoretical) length 3.5 along the negative x axes.
Then draw a "8.2" vector from that endpoint all the way up, 60 degrees from that point (9030: this is because you make a 30 degree angle with the current vertical and you will see that the angle in the triangle we are making is 90 degrees minus that 30).
You should make the 8.2 m vector higher than the starting point of the original 3.5 m vector.
Finally draw a 15 m vector from the endpoint of the 8.2 m vector that goes above the original one, and to the left. This one should be drawn really far.
Now we just solve for the sides. Our first triangle is a 306090, so we must solve for the vertical side (the one with same slope as the 3.5 vector), which is just:
8.2/2 = 4.1 m
Now solve for the opposite side using 4.1(sqrt(3)) = 7.1 m. So now we have enough to solve for the other triangle that the 15m line makes.
the top side is 15  7.1 = 7.9 m
while the right side is 4.1  3.5 = 0.6 m
The last side will be the resultant displacement:
√(7.9^2) + (0.6^2) = 7.9 m (ANSWER)
To find the direction simply use:
arcTan(y/x)=arcTan(0.6/7.9) = 4.3 degrees north of west (ANSWER 2)

Let east be +x and north be +y, x = distance * cos Î¸, y = distance * sin Î¸
For 3.5 south, y = 3.5
For 8.2, x = 8.2 * cos 30 and y = 8.2 * sin 30
For 15 west, x = 15
Total x = 8.2 * cos 30 – 15 â 7.9
Total y = 3.5 + 8.2 * sin 30 = 0.6
Magnitude = [(8.2 * cos 30 – 15)^2 + 0.6^2)]^0.5
The magnitude is approximately 7.9 m
Tan Î¸ = y/x = 0.6/7.9
The angle is approximately 4.3Ë North of west