Question

if the position vector of a particle is given by r = (4cos 2t)i+(4sin2t)j+(6t)k m calculate its acc. at t=x/4

Answer 1

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Given, r = (4cos 2t) i + (4sin2t) j + (6t)k


〽 Since, velocity 【v 】 = dr/dt


So, dr/dt = d/dt [ ( 4cos 2t) i + ( 4sin 2t ) j+ (6t) k]


=> (-8sin2t) i+(8cos2t) j+6k


〽Now, acceleration, 【a】 = dv/dt


dv/dt = d/dt [ ( -8sin 2t ) i + ( 8 cos 2t ) j + 6k ]


=> [ ( -16 cos 2t ) i - ( 16 sin 2t ) j


On putting t = x/4 we get ;


a = [ (-16cos 2x/4) i - (16sin 2x/4) j ]


=> [ (-16cos x/2) i - (16sin x/2)j ]


=> [ -16 (cos x/2 i+sin x/2 j) ]



Hence, acceleration = [ -16 (cos x/2 i+sin x/2 j) ]




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Answer 2

Explanation:

hope it's help you .............