A particle moves along the parabolic path x=y2+2y+2 in such a way that the y− component of velocity vector remains 5 ms−1 during the motion. The magnitude of the acceleration of the particle (in ms−2) is

A particle moves along the parabolic path x=y2+2y+2 in such a way that

1.	A particle moves along the parabolic path x=y2+2y+2 in such a way that the y− component of velocity vector remains 5 ms−1 during the motion. The magnitude of the acceleration of the particle (in ms−2) is
Answer» A particle moves along the parabolic path $x = y^{2} + 2 y + 2$ in such a way that the $y -$ component of velocity vector remains $5 {ms}^{- 1}$ during the motion. The magnitude of the acceleration of the particle (in $m s^{- 2}$ ) 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 5 ms−1 during the motion. The magnitude of the acceleration of the particle (in ms−2) is

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