A particle of mass 2.0×10^2 kg is moving with uniform speed of 4.0 metre per second in circular path of radius 1.0 M find with the help of a vector diagram i) the change in velocity vector B minus vector a of the particle in moving from a to b ii) the change in the centripetal force vector FB - vector FA
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Since the velocities r perpendicular
Delta v=vb-va
=under root(4×4+4×4)
=under root (16+16)
=under root (32)
= 5.7m/s
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