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

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

$4 {ms}^{- 2}$

$\sqrt{5} {ms}^{- 2}$

$2 {ms}^{- 2}$

answer is D.

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Detailed Solution

$Speed, v = \sqrt{v_{x}^{2} + v_{y}^{2}}$

Rate of change of speed,

$\begin{array}{l} \frac{d v}{d t} = \frac{2 v_{x} \frac{d v_{x}}{d t} + 2 v_{y} \frac{d v_{y}}{d t}}{2 \sqrt{v_{x}^{2} + v_{y}^{2}}} = \frac{v_{x} a_{x} + v_{y} a_{y}}{\sqrt{v_{x}^{2} + v_{y}^{2}}} \\ = \frac{(3 \times 2) + (4 \times 1)}{\sqrt{(3)^{2} + (4)^{2}}} = 2 m / s^{2} \end{array}$