Question

The position of a particle moving under uniform accelerationis some function of time and the acceleration. Suppose we writethis position as: x=ka^mt^n where k is adimensionless constant. Show by dimensional analysis that thisexpression is satisfied if m=1 and n=2. Thus bydimensional analysis this expression issatisfied.

Answer 1

For
`m=1` and
`n=2`, the expression is satisfied dimensionally.

Step-by-step explanation:

`x=ka^mt^n`

`x` is a position and has a dimension of length,
`L`.

`a` is acceleration which has dimensions of
`LT^(-2)`.

`t` is time with dimension of
`T`.

Since
`k` is dimensionless, we do not factor it into the dimensional equation as below:

`L = (LT^(-2))^mT^n`

Expanding the first term on the right hand side,

`L = L^mT^(-2m)T^n`

Applying the laws of indices,

`L = L^mT^(-2m+n)`

The index of each fundamental dimension must be equal on both sides.

For
`L`,

`1=m`

For
`T`,

`0=-2m+n`

But
`1=m`

`0=-2*1+n`

`0=-2+n`

`2=n`

Thus, the equation is dimensionally satisfied for the given values of
`m=1` and
`n=2`.