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!
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 (90-30: 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 30-60-90, 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