The relation between time t and displacement x is `t = alpha x^2 + beta x,` where `alpha and beta` are constants. The retardation is

Explanation

`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)`