A particle is moving eastwards with a velocity 5m/s in 10seconds.The velocity changes to 5m/s northwards.Find the magnitude and direction of average acceleration in this time.

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.