A particle undergoes two displacements. The first has a
magnitude of 150 cm and makes an angle of 120° with
the positive x axis. The resultant displacement has a magnitude
of 140 cm and is directed at an angle of 35.0° to
the positive x axis. Find the magnitude and direction of
the second displacement.

Question

A particle undergoes two displacements. The first has a
magnitude of 150 cm and makes an angle of 120° with
the positive x axis. The resultant displacement has a magnitude
of 140 cm and is directed at an angle of 35.0° to
the positive x axis. Find the magnitude and direction of
the second displacement.

goformit100 · Answer

@bahrom7893

bahrom7893 · Answer

god i hate vectors. @.Sam.

.sam. · Answer

Hmm alright I'll think of it since you're going for it too

bahrom7893 · Answer

Ok, this is what I found:
x component of the total final displacement = 140 cos 35 = 114.68
y component of the total final displacement = 140 sin 35 = 80.3

x component of the first displacement vector = -150 x cos 60 = -75
y component of the first displacement vector = 150 x sin 30 = 75

Thus,
x component of the second displacement vector = 114.68 - 75 = 39.68
y component of the second displacement vector = 80.3 - (- 75) = 155.3

Magnitude of the 2nd vector is: sqrt(39.68^2+155.3^2) = 160.28

bahrom7893 · Answer

trying to draw the picture now http://www.twiddla.com/1166652

.sam. · Answer

Erm 
$$Y:150\cos(30)+xsin(\theta)+140\sin(35)=0 \ \ X:150\sin(30)=xcos(\theta)+140\sin(35)$$