Question

Vector A has a magnitude of 4 m/s and it is on the positive x axis .

Vector B has a magnitude of 6 m/s and forms a 30 degrees angle with the positive x axis.

Vector C has a magnitude of 8 m/s and forms a 60 degree angle with the negative x axis.

Find:

a) The magnitude of A + B + C = D

Give your answer with two significant figures.

b) The angle vector D form with the x axis.

Give the answer with two significant figures.

c) What quadrant is vector D?

Answer 1

Suppose given that Vector is R and it makes angle with +x-axis, then it's components are given by:

Rx = R*cos

Ry = R*sin

Now we need to find A + B + C

Using above rule:

A = 4 m/s, direction +ve x-axis

Ax = 4*cos 0 deg = 4 m/sec

Ay = 4*sin 0 deg = 0 m/s

B = 6 m/s, direction 30 deg with +ve x-axis

Bx = 6*cos 30 deg = 5.20 m/s

By = 6*sin 30 deg = 3.00 m/s

C = 8 m/s, direction 60 deg with -ve x-axis

Cx = -8*cos 60 deg = -4.0 m/s

Cy = 8*sin (60 deg) = 6.93 m/s

Now

D = A + B + C

D = (Ax + Bx + Cx) i + (Ay + By + Cy) j

D = (4 + 5.20 - 4) i + (0 + 3.00 - 6.93) j

D = 5.20 i - 3.93 j

So Magnitude of D will be

|D| = sqrt (Rx^2 + Ry^2)

|D| = sqrt (5.20^2 + (-3.93)^2)

|D| = 6.518 m/s = 6.5 m/sec

Direction will be given by:

theta = arctan (Ry/Rx)

theta = arctan ((-3.93)/5.20)

theta = -37.08 deg = -37 deg angle with clockwise with the -ve x axis

Since x > 0 and y < 0, then vector will be in 4th quadrant

Please Upvote.