particle undergoes three displacements. The ?rst has a magnitude of 15 m and mak

ID: 2133611 • Letter: P

Question

particle undergoes three displacements.

The ?rst has a magnitude of 15 m and makes

an angle of 41? with the positive x axis. The

second has a magnitude of 6.3 m and makes

an angle of 154? with the positive x axis. After the third displacement the particle returns

to its initial position.

Find the magnitude of the third displacement.

Answer in units m.

Find the angle of the third displacement (measured from the positive x axis, with counterclockwise positive within the limits of ?180?

to +180

Answer in units of ?

I have the first part right it's the 2nd part that I am getting wrong! PLEASE HELP!

The components of the first are...

x = 15(cos 41) = 11.3

y = (15)(sin 41) = 9.84

The components of the second are

x = (6.4)(cos 26) = -5.75

y = (6.4)(sin 26) = 2.8

Net x = 11.3 - 5.75 = 5.55

Net y = 9.84 + 2.8 = 12.64

Net vector is sqrt[(5.55)^2 + (12.64)^2)]

Net = 13.8m

The angle is from tan(angle) = 12.64/5.55

angle = 66.3 degrees

That is -113.7 degrees from the positive x axis

(NOTE: this can also be expresses as 246.3 degrees) Although your answer key wants is at -113.7

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.