**Solution: We have**

Initial velocity, u = 5 m/s due east

Final velocity, v = 5 m/s due north

Change in velocity, Δv = v – u = 5 m/s due north - 5 m/s due east

=> Δv = 5 m/s due north + 5 m/s due west [negative of east direction is west]

Δv = (v^{2} + u^{2})^{1/2}

=> Δv = 5√2 m/s

Direction of Δv is given by,

tanθ = v/u = -1

=> θ = -45^{o}

**Thus, the acceleration is a = Δv/t = 5√2/10 = 1/√2 m/s**^{2} and is directed towards North-West direction.