Velocity (v) of a particle moving along x -axis varies with its position x as shown in figure. accel

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2.Motion in Straight Line

hard

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

$a=4 x-8$

Solution

(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 $x$

$\Rightarrow \frac{v-4}{x-0}=\frac{0-4}{2-0}$

$\Rightarrow \frac{v-u}{x}=\frac{-4}{2}$

$\Rightarrow v-4=-2 x$

$\Rightarrow v=-2 x+4$

$\Rightarrow \frac{d v}{d x}=-2$

$\Rightarrow a=\frac{v d v}{d x}=(-2 x+4)(-2)$

$\Rightarrow a=4 x-8$

Standard 11

Physics

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