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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). |
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Answer» `sqrt2` `(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)` |
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