A scooter going due east at 10 ms-1 turns right through an angle of 900. If the speed of the scooter remains unchanged in taking turn, the change in the velocity of the scooter is

If the magnitude of vector remains same, only direction change by θ thenΔv→ = v2→ −v1,→Δv→ = v2→ +(−v1→)Magnitude of change in vector |Δv→|   =  2vsinθ2|Δv→| = 2×10×sin(9002) = 102  =  14.14 m/sDirection is south-west as shown in figure.