Question

A vector has the components Ax = -33 m and Ay = 44 m. What is the magnitude of this vector? What angle does this vector make with the positive x-axis?

Answer 1

Step-by-step explanation:

To find the magnitude of the vector, we can use the Pythagorean theorem:

|A| = sqrt(Ax^2 + Ay^2)

where Ax and Ay are the components of the vector A.

Substituting the given values:

|A| = sqrt((-33 m)^2 + (44 m)^2)

|A| = sqrt(1089 m^2 + 1936 m^2)

|A| = sqrt(3025 m^2)

|A| = 55 m

Therefore, the magnitude of the vector is 55 m.

To find the angle that the vector makes with the positive x-axis, we can use the inverse tangent function:

theta = tan^-1(Ay / Ax)

Substituting the given values:

theta = tan^-1(44 m / -33 m)

theta ≈ -53.13°

Note that the negative sign indicates that the angle is measured clockwise from the positive x-axis. If we want the angle to be measured counterclockwise from the positive x-axis, we can add 360°:

theta = 360° + (-53.13°)

theta ≈ 306.87°

Therefore, the angle that the vector makes with the positive x-axis is approximately 306.87°.