Square of velocity and position graph of a particle is given, at point A slope is 45° then acceleration of particle at ,A is :

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0.5 m/s²

4 m/s²

1 m/s²

2 m/s²

answer is D.

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Detailed Solution

$\begin{array}{l} Slope of v^{2} - x graph = slope = \frac{{dv}^{2}}{dx} = 2 v \frac{dv}{dx} = 2 a \\ \Rightarrow \tan 45^{\circ} = 2 a \Rightarrow a = \frac{1}{2} m / s^{2} \end{array}$

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