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

Question

The velocitydisplacement $(v-s)$ graph shows the motion of particle moving in a straight linè.

Velocity-displacement graph is a circle of radius $2 \mathrm{~m}$ and centre is at $(2,0) \mathrm{m}$. The value of acceleration for this particle at a point $(2-\sqrt{2}, \sqrt{2}) \mathrm{m}$ will be $\mathrm{ms}^{-2}$.

Aicharya · Accepted Answer

As, graph is a circle with centre $(2,0)$ and radius 2 , its equation is $ \begin{aligned} (s-2)^2+(v-0)^2 & =2^2 \ (s-2)^2+v^2 & =4 \end{aligned} $ Differentiating with respect to time, we get $ \begin{aligned} 2(s-2) \frac{d s}{d t}+2 v \frac{d v}{d t} & =0 \ 2(s-2) v+2 v \cdot a & =0 \end{aligned} $ At, $s=2-\sqrt{2}$ and $v=\sqrt{2}$ $ \begin{aligned} & 2(2-\sqrt{2}-2) 	imes \sqrt{2}+2 \sqrt{2} a=0 \ \Rightarrow & -4+2 \sqrt{2} a=0 \Rightarrow a=\frac{4}{2 \sqrt{2}}=\sqrt{2} \mathrm{~ms}^{-2} \end{aligned} $

The velocitydisplacement $(v-s)$ graph shows the motion of particle moving in a straight linè.…

Solution