A vector A has components Ax = −4.00 m and Ay = 3.50 m. Find the magnitude (in m) and the direction (in degrees counterclockwise from the +x-axis) of the vector.

Part (i) the magnitude of the vector is 5.315 m

Part (ii) the direction of the vector is 138.8⁰

Explanation:

Given;

y-component Ay = 3.50 m

x -component Ax = -4.00 m

Vector representation = (-4i + 3.5j)

Part (i) the magnitude of the vector

Magnitude of the vector = √(-4)² +(3.5)²

Magnitude of the vector = √28.25

                                          = 5.315 m

Part (ii) the direction (in degrees counterclockwise from the +x-axis) of the vector.

If we make a rough sketch of this vector, the direction of this vector lies in second quadrant.

That is; 90 < θ < 180

Let's solve for θ

tan θ = y/x

tan θ = 3.5/-4

tan θ = - 0.875

θ = tan⁻¹ (-0.875)

θ = - 41.2⁰

Since θ lies in the second quadrant,

θ = - 41.2⁰ = 180 - 41.2 = 138.8⁰

The direction of the vector is 138.8⁰

true tan speed = ang vel x radius