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, parabolic path x=y 2 + 2y + 2 differentiate w,r,t, t dx/dt = 2y dy/dt + 2 dy/dt Vx = 2 y Vy + 2 Vy Vx = 10y + 10 , again differentiating Ax = 10 dy/dt Ax = 50 m/sec^2 total Accl. = &radic; (Ax)^2 + (Ay)^2 Ay = 0 & Ax = 50 m/ sec^2 total accl. = 50 m/sec^2

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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=y² + 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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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=y2 + 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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A particle mover along the parabolic path x=y² + 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-