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

Answer 1

Answer: (D)

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