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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 = (v2 + u2)1/2

=> Δv = 5√2 m/s

Direction of Δv is given by,

tanθ = v/u = -1

=> θ = -45o

Thus, the acceleration is a = Δv/t =  5√2/10 = 1/√2 m/s2 and is directed towards North-West direction.

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