Question

Find out the magnitude as well as the direction of the following vectors:

A_x = 22.0 cm; A_y = 33.0 cm;

B_x = -11.0 m; B_y = 84.5 m;

C_x = -77.3 mm; C_y = -66.7 mm;

D_x = 34.5 m/s; D_y = -56.9 m/s.

Answer 1

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

Rx = R*cos

Ry = R*sin

Now to find R and ,

R = sqrt (Rx^2 + Ry^2)

= arctan (Ry/Rx)

Also 1st quadrant both x and y +ve, In 2nd quadrant x is -ve and y is +ve, In 3rd quadrant both x and y is

-ve, In 4th quadrant x is +ve and y is -ve.

Using above rule:

1.

Ax = 22.0 cm; Ay = 33.0 cm;

A = sqrt (22^2 + 33^2) = 39.66 cm

Angle = arctan (33/22) = 56.31 deg CCW with +ve x-axis

2.

Bx = -11.0 m; By = 84.5 m

B = sqrt ((-11)^2 + 84.5^2) = 85.21 m

Angle = arctan (84.5/11) = 82.58 deg CW with -ve x-axis

Angle = 180 - 82.58 = 97.42 deg CCW with +ve x-axis

C.

Cx = -77.3 mm; Cy = -66.7 mm

C = sqrt ((-77.3)^2 + (-66.7)^2) = 102.1 mm

Angle = arctan (66.7/77.3) = 40.8 deg CCW with -ve x-axis

Angle = 180 + 40.8 = 220.8 deg CCW with +ve x-axis

D.

Dx = 34.5 m/s; Dy = -56.9 m/s.

C = sqrt ((34.5)^2 + (-56.9)^2) = 66.54 m/s

Angle = arctan (56.9/34.5) = 58.77 deg CW with +ve x-axis

Angle = 360 - 58.77 = 301.23 deg CCW with +ve x-axis

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