A particle is moving along $x$ axis and its velocity (v) vs position $(x)$ graph is a curve as shown in the figure. Line $A P B$ is normal to the curve at point $P$. Find the instantaneous acceleration of the particle at $x=3.0 \mathrm{~m}$.

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Aicharya · Accepted Answer

Let the velocity of the particle at $x=3.0 \mathrm{~m}$ be $v_{0}$. The slope of line $A P B=-\frac{v_{0}}{1}$ As the line $A P B$ is normal to the curve at point $P$, hence the slope of tangent at $P=\frac{1}{v_{0}}$ It means the slope of $v-x \operatorname{graph}\left(\frac{d v}{d x}\right)_{P}=\frac{1}{v_{0}}$ ...(i) But we define acceleration as $a=v \frac{d v}{d x}$ ...(ii) But at $x=3.0 \mathrm{~m}, v=v_{0}$ and $\frac{d v}{d x}=\left(\frac{1}{v_{0}}\right) a=v_{0}\left(\frac{1}{v_{0}}\right)=1 \mathrm{~m} / \mathrm{s}^{2}$ Hence $a=v_{0}\left(\frac{1}{v_{0}}\right)=1 \mathrm{~m} / \mathrm{s}^{2}$

A particle is moving along \(x\) axis and its velocity (v) vs position \((x)\) graph is a curve as shown in…

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