The velocity (v) of a particle moving along t | vedclass

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(D) The acceleration $a$ is given by the formula $a = v \frac{dv}{dx}$.First,we find the equation of the line representing velocity $v$ as a function

Vedclass · Accepted Answer

$a = 4x - 8$

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(D) The acceleration $a$ is given by the formula $a = v \frac{dv}{dx}$.First,we find the equation of the line representing velocity $v$ as a function of position $x$. The line passes through the points $(0, 4)$ and $(2, 0)$.The slope $m$ of the line is given by $m = \frac{v_2 - v_1}{x_2 - x_1} = \frac{0 - 4}{2 - 0} = \frac{-4}{2} = -2$.The $y$-intercept $c$ is $4$.Thus,the equation of the line is $v = mx + c$,which gives $v = -2x + 4$.Now,we differentiate $v$ with respect to $x$ to find $\frac{dv}{dx}$:$\frac{dv}{dx} = \frac{d}{dx}(-2x + 4) = -2$.Finally,we substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:$a = v \frac{dv}{dx} = (-2x + 4)(-2) = 4x - 8$.Therefore,the correct option is $(d)$.

The velocity $(v)$ of a particle moving along the $x$-axis varies with its position $(x)$ as shown in the figure. How does the acceleration $(a)$ of the particle vary with position $(x)$?

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