The velocity - displacement (v-s) graph shows the motion of particle moving in a straight line.
Velocity-displacement graph is a circle of radius 2 m and centre is at (2, 0) m.
The value of acceleration for this particle at a point `(2-sqrt2,sqrt2)`m will be .............. `ms^(-2)`.

Options

A. `sqrt2`
B. 4
C. 2
D. 3

Answer

A

Explanation

As, graph is a circle with centre (2, 0) and radius 2, its equation is
`(s-2)^(2)+(v-0)^(2)=2^(2)`
`(s-2)^(2)+v^(4)=4`
Differentiating with respect to time, we get
`2(s-2)(ds)/(dt)+2v(dv)/(dt)=0`
`2(s-2)v+2v*a=0`
At, `s=2-sqrt2andv=sqrt2`
`2(2-sqrt2-2)xxsqrt2+2sqrt2a=0`
`rArr-4+2sqrt2a=0rArra=4/(2sqrt2)=sqrt2ms^(-2)`