A particle is moving in the x-y plane. At certain instant of time,the components of its velocity and acceleration are as follows: $v_{x} = 3 m s^{- 1}$ , $v_{y} = 4 m s^{- 1}$ , $a_{x} = 2 m s^{- 2}$ and $a_{y} = 1 m s^{- 1}$ . The rate of change of speed at this moment is:

\sqrt{10} m s^{- 2}

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4 m s^{- 2}

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\sqrt{5} m s^{- 2}

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2 m s^{- 2}

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Solution

The correct option is A $\sqrt{5} m s^{- 2}$
Given,

$V_{x} = 3 m / s, a_{x} = 2 m / s^{2}, V_{y} = 4 m / s, a_{y} = 1 m / s^{2}$

So, $a_{r e s u l t a n t} = \sqrt{a_{x}^{2} + a_{y}^{2}} = \sqrt{4 + 1} = \sqrt{5} m / s^{2}$ in x-y direction

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x - y

v_{x} = 3 {ms}^{- 1}, v_{y} = 4 {ms}^{- 1}, a_{x} = 2 {ms}^{- 2}

a_{y} = 1 {ms}^{- 2}

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