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

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pratham ashish · Answer

Hi, A = root{(Ax)^2 + (Ay)^2} A= total acceleration Ax = acceleration along x axis = d^2x/dt^2 Ay = acceleration along y axis = d^2y/dt^2 here vel. in y directn =dy/dt = 5 m/s (constant) => Ay = 0 now differentiate the given eqn => Vx = 2y*dy/dt +2* dy/dt = 2y*5 + 2 *5 Vx = 10y + 10 again diff. Ax = 10 *dy/dt = 10* 5 = 50 m/s^2 A = root{(Ax)^2 + (Ay)^2} => A = Ax = 50 (as Ay=0 )

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

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

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