Question

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

• A. $a^2=x+3$
• B. $a=2x^2+4$
• C. $2a=3x+5$
• D. $a=4x-8$

Answer 1

Answer: (D)

$a=\frac{vdv}{dx}$
$\frac{dv}{dx}\to$ slope
So, $\frac{-4}{2}=-2$
Intercept $=+4$
$a\to$ Negative
So, $a=\frac{vdv}{dx}$
Relation between $v$ and $x$
$\Rightarrow \frac{v-4}{x-0}=\frac{0-4}{2-0}$
$\Rightarrow \frac{v-u}{x}=\frac{-4}{2}$
$\Rightarrow v-4=-2x$
$\Rightarrow v=-2x+4$
$\Rightarrow \frac{dv}{dx}=-2$
$\Rightarrow a=\frac{vdv}{dx}=(-2x+4)(-2)$
$\Rightarrow a=4x-8$