Set 5 Problem number 6

What are x and y the components of the velocity vector obtained when we add the two following velocity vectors:

vector A, with x and y components -2 m/s and -6.9 m/s, and
vector B whose x and y components are -8.3 m/s and -4.1 m/s?

What are the magnitude and angle of the resultant vector?

Solution

The x component is simply the sum -2 m/s + -8.3 m/s = -10.31 m/s of the x components of the vectors being added.

The y component is similarly the sum -6.9 m/s + -4.1 m/s = -11 m/s of the y components of the added vectors.

The magnitude of the resultant vector, by the Pythagorean Theorem, is therefore

magnitude = `sqrt( ( -10.31 m/s )² + ( -11 m/s )²) = 15.07 m/s .

The angle of the resultant vector to the x axis is

angle = tan^-1( -11 /( -10.31 )) = 46.85 degrees or 46.85 degrees + 180 degrees.

Since the x component of this vector is negative, the vector is in the second quadrant and the correct angle will be

angle = ( 46.85 + 180) degrees = 226.85 degrees.

Vectors A and B, with their x and y components indicated, are shown on the figure below. The sum of the two x components will be the x component of the resultant, and the sum of the two y components will be the y component of the resultant:

R_x = A_x + B_x

and

R_y = A_y + B_y.

We find the magnitude and angle of R, using the Pythagorean Theorem and the inverse tangent:

| R | = `sqrt(R_x² + R_y²)

and

`theta = tan^-1(R_y / R_x).

We understand that if R_x is negative, we must add 180 degrees to the inverse tangent in order to place the angle in either the third or fourth quadrant.

Explanation in terms of Figure(s), Extension

The figure below shows two vectors A and B, and their components A_x, A_y, B_x and B_y.

The y components A_y and B_y are seen to add up to R_y, as the x components A_x and B_x are seen to at up to R_x.
The magnitude and angle of the vector R are obtained by using the Pythagorean Theorem and the inverse tangent in the usual manner.