The given graph shows the variation of veloci | vedclass

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(B) The given graph is a straight line with a positive intercept $v_0$ and a negative slope. The equation of the line is given by:$v = -mx + v_0$ ...

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(B) The given graph is a straight line with a positive intercept $v_0$ and a negative slope. The equation of the line is given by:$v = -mx + v_0$ ... $(i)$where $m = 	an 	heta = \frac{v_0}{x_0}$ is the magnitude of the slope.We know that acceleration $a = v \frac{dv}{dx}$.From equation $(i)$,differentiating with respect to $x$ gives:$\frac{dv}{dx} = -m$Substituting this into the acceleration formula:$a = (-mx + v_0)(-m)$$a = m^2x - mv_0$This is the equation of a straight line with a positive slope $(m^2)$ and a negative intercept $(-mv_0)$ on the $a$-axis.Comparing this with the given options,Graph $B$ represents a straight line with a negative intercept on the $a$-axis and a positive slope. Therefore,Graph $B$ is correct.

The given graph shows the variation of velocity $v$ with displacement $x$. Which one of the graphs given below correctly represents the variation of acceleration $a$ with displacement $x$?

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