The velocity-displacement (v-s) graph shows t | vedclass

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(A) The graph is a circle with center $(2, 0)$ and radius $2 \ m$. Its equation is given by:$(s-2)^2 + v^2 = 2^2$$(s-2)^2 + v^2 = 4$Differentiating bo

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$\sqrt{2}$

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(A) The graph is a circle with center $(2, 0)$ and radius $2 \ m$. Its equation is given by:$(s-2)^2 + v^2 = 2^2$$(s-2)^2 + v^2 = 4$Differentiating both sides with respect to time $t$,we get:$2(s-2) \frac{ds}{dt} + 2v \frac{dv}{dt} = 0$Since $\frac{ds}{dt} = v$ and $\frac{dv}{dt} = a$,we substitute these into the equation:$2(s-2)v + 2v \cdot a = 0$Dividing by $2v$ (assuming $v 
eq 0$):$(s-2) + a = 0$$a = -(s-2) = 2-s$At the point $(s, v) = (2-\sqrt{2}, \sqrt{2})$:$a = 2 - (2-\sqrt{2}) = \sqrt{2} \ ms^{-2}$

The velocity-displacement $(v-s)$ graph shows the motion of a particle moving in a straight line. The velocity-displacement graph is a circle of radius $2 \ m$ and the center is at $(2, 0) \ m$. The value of acceleration for this particle at a point $(2-\sqrt{2}, \sqrt{2}) \ m$ will be $ms^{-2}$.

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