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

Answer:

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

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 $1=m$

$0=-2\times1+n$

$0=-2+n$

$2=n$

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