The given graph shows the variation of veloci | vedclass

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(A) From the given $v-x$ graph,the equation of the straight line is $v = mx + c$. Here,the intercept $c = 20$ and the slope $m = \frac{0 - 20}{10 - 0}

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(A) From the given $v-x$ graph,the equation of the straight line is $v = mx + c$. Here,the intercept $c = 20$ and the slope $m = \frac{0 - 20}{10 - 0} = -2$. So,the equation is $v = -2x + 20$. We know that acceleration $a = v \frac{dv}{dx}$. First,find $\frac{dv}{dx} = \frac{d}{dx}(-2x + 20) = -2$. Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula: $a = (-2x + 20)(-2) = 4x - 40$. This is the equation of a straight line with a slope of $4$ and a y-intercept of $-40$. At $x = 0$,$a = -40$. At $x = 10$,$a = 4(10) - 40 = 0$. Thus,the graph represents a straight line starting from $a = -40$ at $x = 0$ and reaching $a = 0$ at $x = 10$. Therefore,option $A$ is correct.

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

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