Question

During the rodeo, a clown runs 11m north, turns 24º east of north, and runs 2.1m. Then, after waiting for the bull to come near, the clown turns due east and runs 8.3, to exit the arena.
What is the magnitude of the clown's total displacement?

How many degrees east of north is the clown's total displacement?

Answer 1

D=11N+2.1*cos24 N +2.1sin24E + 8.3E
add like terms.
displacement= sqrt (xxx^2 N + yyyy^2E) you figure xxxx and yyyy from above.
angle=arctan yyyy/xxxxx E of N

Answer 2

All angles are measured CW from +y-axis.

Disp. = 11m[0o] + 2.1m[24o] + 8.3m[90o].
X = 11*sin 0 + 2.1*sin24 + 8.3*sin90 = 9.15 m.
Y = 11*Cos 0 + 2.1*Cos24 + 8.3*Cos90 = 12.92 m.

Disp. = X + Yi = 9.15 + 12.92i = 15.83m[35.3o] N. of E.
Tan A = X/Y.
A = 35.3o

Answer 3

Correction: A = 35.3o E. of N.