Question

the velocity vector of a car has an x- component of 17m/sec and a y-component of -11m/sec. What are the magnitude and direction of the velocity vector?

Answer 1

The velocity vector is:

$$\mathbf{v} = (v_x,v_y) = (17,-11)\space m/s$$

The magnitude is:

$$v = \sqrt{v_x^2 + v_y^2 } = \sqrt{17^2 + (-11)^2 } \approx 20.25\space m/s$$

The directon is:

$$\theta = \arctan \left( \dfrac{v_y}{v_x} \right) \approx 327.1$$

Answer. $v = 20.25\space m/s$ and $\theta = 327.1$ counting from the x-axis.