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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 θ made with the positive horizontal axis, satisfying
tan(θ) = (4 m/s) / (-3 m/s) = -4/3 → θ = arctan(-4/3) + 180°n
where n is any integer. Now you have to consider that the x coordinate is negative and the y coordinate is positive, so 〈-3, 4〉 points into the second quadrant, and we get an angle there for n = 1 of about
θ ≈ 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 θ is such that
〈-3, 4〉 • 〈1, 0〉 = |〈-3, 4〉| |〈1, 0〉| cos(θ)
(-3)•1 + 4•0 = √((-3)² + 4²) • √(1² + 0²) cos(θ)
-3/5 = cos(θ) → θ = arccos(-3/5) + 360°n
where again, n is any integer, and we get the same solution θ ≈ 126.87° in the second quadrant when n = 0.