A particle moves with a constant speed v in a circular path. When the particle covers an angular displacement of 60^ , what will be the magnitude of change in its velocity and the change in magnitude of its velocity, respectively?

Step-by-step solution

Let the initial velocity be v ₁ and final velocity be v ₂ . Since the particle moves with a constant speed v , the magnitude of velocity remains constant. Therefore, the change in magnitude of velocity is: | v | = | v ₂| - | v ₁| = v - v = 0 The magnitude of change in velocity is given by the vector subtraction of v ₁ from v ₂ : | v | = | v ₂ - v ₁| = v₁^2 + v₂^2 - 2v₁v₂ Given = 60^ and v₁ = v₂ = v : | v | = v^2 + v^2 - 2(v)(v) 60^ | v | = 2v^2 - 2v^2 ( 1 2 ) = v^2 = v Thus, the magnitude of change in velocity is v