If the time and displacement of the particle along the positive x-axis are related as `t=(x^(2)-1)^(1//2)`, find the acceleration in terms of `x`.

Explanation

`t=(x^(2)-1)^(1//2)`
`(dt)/(dx)=1/2(x^(2)-1)^(1//2)(2x)=x/((x^(2)-1)^(1//2))`
`v=((x^(2)-1)^(1//2))/x=(1-1/x^(2))^(1//2)`
`(dv)/(dx)=1/2(1-1/x^(2))^(-1//2)(2/x^(3))`
`a=v(dv)/(dx)=1/x^(3)`
OR
`t=(x^(2)-1)^(1//2)impliesx=(1+t^(2))^(1//2)`
`v=(dx)/(dt)=1/2(1+t^(2))^(-1//2)(2t)`
`=t/((1+t^(2))^(1//2))=1/((1/t^(2)+1)^(1//2))=(1+1/t^(2))^(-1//2)`
`a=(dv)/(dt)=-1/2(1+1/t^(2))^(-3//2)(-2/t^(3))`
`=1/(t^(3)(1+1/t^(2))^(3//2))=1/((t^(2)+1)^(3))`
`=1/x^(3)`
OR
`t=(x^(2) -1)^(1//2)`
`t^(2)=(x^(2)-1)`
Differentiating w.r.t. t, we get
`2t=2x(dx)/(dt)=2xv`
`t=xv`
Differentiating w.r.t. t we get
`1=x(dv)/(dt)+v(dx)/(dt)=xa+v^(2)` (Product rule)
`ax=1-v^(2)=1-(t^(2))/(x^(2))=(x^(2)-t^(2))/(x^(2))=(1)/(x^(2))`
`a=(1)/(x^(3))`
Though the above method is simple, but for this, you will require further knowledge of differentiation. e.g. if x and y are functions of `t` (not shown in expression),
`y=x^(3)+3x^(2)`
differentiating w.r.t. t, we get
`(dy)/(dt)=3x^(2)(dx)/(dt)+3.2x(dx)/(dt)`
Differentiating w.r.t. `t`, we get
`2y(dy)/(dt)=3x^(2)(dx)/(dt)`
We can also write
`2y dy = 3x^(2) dx`