The given graph shows the variation of veloci | vedclass

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(A) The given graph is a straight line with a positive intercept on the $v$-axis and a negative slope.Its equation can be written as:$v = -mx + v_0 \d

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(A) The given graph is a straight line with a positive intercept on the $v$-axis and a negative slope.Its equation can be written as:$v = -mx + v_0 \dots(1)$where $m = 	an 	heta = \frac{v_0}{x_0}$ is the magnitude of the slope.The acceleration $a$ is given by $a = v \frac{dv}{dx}$.From equation $(1)$,differentiating $v$ with respect to $x$ gives:$\frac{dv}{dx} = -m$Substituting this into the expression for acceleration:$a = v(-m) = (-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 acceleration axis.Comparing this with the given options,graph $(A)$ represents a straight line with a negative intercept and a positive slope.Therefore,option $(A)$ is correct.

The given graph shows the variation of velocity $(v)$ with position $(x)$ for a particle moving along a straight line. Which of the following graphs shows the variation of acceleration $(a)$ with position $(x)$?

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