The acceleration of a particle which moves along the positive $x$-axis varies with its position as shown in the figure. If the velocity of the particle is $0.8 \mathrm{~ms}^{-1}$ at $x=0$, then its velocity at $x=1.4 \mathrm{m}$ is (in $\mathrm{ms}^{-1}$)

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext {For a - } \mathrm{x} 	ext { graph } \ & \mathrm{v}^2=\mathrm{u}^2+\int_{\mathrm{x}_1}^{\mathrm{x}_2} \mathrm{adx}=\mathrm{u}^2+2 	ext { (Area of a-x graph) } \ & \Rightarrow \mathrm{v}^2=(0.8)^2+2\left[0.4 	imes 0.4+\frac{1}{2}(0.4+0.2) 	imes 0.4+0.2ight. \ & 	imes 0.6] \ & \Rightarrow \mathrm{v}^2=1.44 \ & 	herefore \mathrm{v}=1.2 \mathrm{~m} / \mathrm{s}\end{aligned}$

The acceleration of a particle which moves along the positive $x$-axis varies with its position as shown in…

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