InterviewSolution
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InterviewSolution
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The relation between time t and displacement x is `t = alpha x^2 + beta x,` where `alpha and beta` are constants. The retardation is |
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Answer» `t=alphax^(2)+betax` Differentiating w.r.t. x, we get `(dt)/(dx)=alpha.2x+beta.x` `(1)/v = 2alphax+beta` `v^(-1)=2alphax+beta` Differentiating w.r.t. t, we get `(-1)v^(-2)(dv)/(dt)=2alpha(dx)/(dt)` `-a/v^(2)=2alphav` `a=-2alphav^(3)` `a=-2alphav^(3)` OR `t=alphax^(2)+betax` Differentiating w.r.t. t, we get `1= alpha.2x(dx)/(dt)+beta(dx)/(dt)` `=(2alpha x+beta)v` Differentiating w.r.t. t, we get `0=(2alphax+beta)(dv)/(dt)+(2alpha(dx)/(dt)+0)v` `=(2alphax+beta)a+2alphav^(2)` `a= -(2alphav^(2))/((2alphax+beta))= - 2alphav^(3)` OR `t=alphax^(2)+betax` `alphax^(2)+betax-t=0` `x = (-beta+-(beta^(2)+4alphat)^(1//2))/(2alpha)` `=-(beta)/(2alpha)+(1)/(2alpha)(beta^(2)+4alphat)^(1//2)` `v=(dx)/(dt)=(1)/(2alpha).(1)/(2)(beta^(2)+4alphat)^(1//2-1)(4alpha)` `=(beta^(2)+4alphat)^(-1//2)` `a=(dv)/(dt)=-(1)/(2)(beta^(2) +4alphat)^(-1//2-1)(4alpha)` `=-2alpha(beta^(2)+4alphat)^(-3//2)` `=-2alphap(beta^(2)+4alphat)^(-1//2)]^(3)` `=-2alphav^(3)` |
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