Position of a body with acceleration 'a' is given by $x=K{a}^{m}{t}^{n},$here t is time. Find dimension of m and n

(1) m = 1, n = 1

(2) m = 1, n = 2

(3) m = 2, n = 1

(4) m = 2, n = 2

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(2) As $x=K{a}^{m}×{t}^{n}$

$\left[{M}^{0}L{T}^{0}\right]={\left[L{T}^{-2}\right]}^{m}{\left[T\right]}^{n}=\left[{L}^{m}{T}^{-2m+n}\right]$

$\therefore$ m = 1 and $-2m+n=0$ $⇒n=2$.

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