A particle moves along the parabolic path x=y2 + 2y + 2 in such a way that the y component of velocity vector remains 5m/s during the motion. The magnitude of the acceleration of the particle is?

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Khimraj · Answer

x = y 2 + 2y +2 differentiating wrt time v x = 2yv y + 2v y again differentiating wrt time a x = 2v y 2 + 2ya y + 2a y a y = 0 as v y = 5m/s(constant) then a x = 2*5 2 = 50 m/s 2 So a = = 50m/s 2 Hope it clears.

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A particle moves along the parabolic path x=y2 + 2y + 2 in such a way that the y component of velocity vector remains 5m/s during the motion. The magnitude of the acceleration of the particle is?

A particle moves along the parabolic path x=y2 + 2y + 2 in such a way that the y component of velocity vector remains 5m/s during the motion. The magnitude of the acceleration of the particle is?

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A particle moves along the parabolic path x=y2 + 2y + 2 in such a way that the y component of velocity vector remains 5m/s during the motion. The magnitude of the acceleration of the particle is?

A particle moves along the parabolic path x=y2 + 2y + 2 in such a way that the y component of velocity vector remains 5m/s during the motion. The magnitude of the acceleration of the particle is?

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