His displacement is √(9² + 12²) = √(81 + 144) = √(225) = 15 m
Displacement/time = 15m/5s = 3 m/s
The direction from start-point to end-point is
90° + tan⁻¹(-3/4) = 90° + tan⁻¹(-3/4) = 126.7°
The student's average velocity is
3 m/s in the direction 36.7° west of north.