Velocity is a vector, so it has a magnitude and direction.
Speed is its magnitude, which is obtained from
|〈-3, 4〉 m/s| = √((-3 m/s)² + (4 m/s)²) = √(25 m²/s²) = 5 m/s
Its direction is an angle <em>θ</em> made with the positive horizontal axis, satisfying
tan(<em>θ</em>) = (4 m/s) / (-3 m/s) = -4/3 → <em>θ</em> = arctan(-4/3) + 180°<em>n</em>
where <em>n</em> is any integer. Now you have to consider that the <em>x</em> coordinate is negative and the <em>y</em> coordinate is positive, so 〈-3, 4〉 points into the second quadrant, and we get an angle there for <em>n</em> = 1 of about
<em>θ</em> ≈ 126.87°
Alternatively, you can use the vector 〈1, 0〉 in place of the axis, then compute the angle by relating it to the dot product, so <em>θ</em> is such that
〈-3, 4〉 • 〈1, 0〉 = |〈-3, 4〉| |〈1, 0〉| cos(<em>θ</em>)
(-3)•1 + 4•0 = √((-3)² + 4²) • √(1² + 0²) cos(<em>θ</em>)
-3/5 = cos(<em>θ</em>) → <em>θ</em> = arccos(-3/5) + 360°<em>n</em>
where again, <em>n</em> is any integer, and we get the same solution <em>θ</em> ≈ 126.87° in the second quadrant when <em>n</em> = 0.
