The acceleration of a particle which moves al | vedclass

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Question

$a=v \frac{d v}{d t} \Rightarrow \int_{u}^{v} v d v=\int_{0}^{1.4} a d x$

$\Rightarrow \frac{v^{2}-u^{2}}{2}=$ Areaofa $-x \operatorname{graph}$

Vedclass · Accepted Answer

$1.2$

Vedclass · Answer

$a=v \frac{d v}{d t} \Rightarrow \int_{u}^{v} v d v=\int_{0}^{1.4} a d x$

$\Rightarrow \frac{v^{2}-u^{2}}{2}=$ Areaofa $-x \operatorname{graph}$

$\Rightarrow v^{2}=(0.8)^{2}=2(0.4) \Rightarrow v=1.2$

The acceleration of a particle which moves along the positive $x-$axis varies with its position as shown. If the velocity of the particle is $0.8 m/s$ at $x = 0$ , the velocity of the particle at $x = 1.4$ is(in $m/s$)

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