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?
Code: Select all
Correct Answer: 13 m at 57 degrees North of Eastx1 = 0 m
y1 = 8.0 m
x2 = [cos(55d)](3.5m) = 2 m
y2 = [sin(55d)](3.5m) = 2.9 m
total delta x = 2 m + 5 m = 7 m
total delta y = 10.9 m
c^2 = a^2 + b^2 = (7^2) + (10.9^2) = 49 + 118.81 = 167.81
c (hypotenuse) = 13 m (Ding! Thanks andrew)
tan theta (angle) = opposite / adjacent
theta (angle) = (opposite / adjacent) / tan
theta (angle) = ( 7 m / 10.9 m) / tan = 57 (Andrew, you rock.)
If anyone knows what I'm doing wrong (apparently a lot), feel free to post and tell me how stupid I am, haha.
EDIT: Fixed, thanks so much andrew.
