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Q.

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

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a

$100 {ms}^{- 2}$

b

$0.1 {ms}^{- 2}$

c

$10 \sqrt{2} {ms}^{- 2}$

d

$50 {ms}^{- 2}$

answer is A.

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Detailed Solution

$v_{y} = \frac{d y}{d t} = 5 {ms}^{- 1} (given) ∴ a_{y} = \frac{d v_{y}}{d t} = 0 x = y^{2} + 2 y + 2 (given) ∴ v_{x} = \frac{d x}{d t} = (2 y + 2) \frac{d y}{d t} = 10 (y + 1) a_{x} = \frac{d v_{x}}{d t} = 10 \frac{d y}{d t} = 50 {ms}^{- 2} ∴ a = a_{x} = 50 {ms}^{- 2} (∵ a_{y} = 0)$

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