Question

The displacement of a particle is related with time as `x^2=t^2+2t+3`. How acceleration is related with displacement?

Answer 1

Given that the displacement (`x`) of a particle is related with time `t` as
`x^2=t^2+2t+3.......[1]`

Differentiating w r to `t` we get

`=>2x(dx)/(dt)=2t+2`

`=>x(dx)/(dt)=t+1.....[2]`

Again differentiating w r to `t` we get

`=>x(d^2x)/(dt^2)+[(dx)/(dt)]^2=1`

Multiplying both sides by `x^2`

`=>x^3(d^2x)/(dt^2)+x^2[(dx)/(dt)]^2=x^2`

Inserting `x(dx)/(dt)=t+1` from [2]

`=>x^3(d^2x)/(dt^2)+(t+1)^2=x^2`

Inserting `x^2=t^2+2t+3` from [2]

`=>x^3(d^2x)/(dt^2)+(t^2+2t+1)=t^2+2t+3`

`=>x^3(d^2x)/(dt^2)=2`

`=>(d^2x)/(dt^2)=2/x^3`

Hence the required relation

`color(red)("Acceleration"(a)=2/{"Displacement"(x)]^3)`