Position of a body with acceleration 'a&# | vedclass

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Question

(b) As $x = K{a^m} 	imes {t^n}$

$\left[ {{M^0}L{T^0}} ight] = {\left[ {L{T^{ - 2}}} ight]^m}{\left[ T ight]^n} = \left[ {{L^m}{T^{ - 2m + n}

Vedclass · Accepted Answer

$m = 1,\;n = 2$

Vedclass · Answer

(b) As $x = K{a^m} 	imes {t^n}$

$\left[ {{M^0}L{T^0}} ight] = {\left[ {L{T^{ - 2}}} ight]^m}{\left[ T ight]^n} = \left[ {{L^m}{T^{ - 2m + n}}} ight]$

$	herefore $ $m = 1$ and $ - 2m + n = 0$  $&rArr;$  $n = 2$.

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 $n$.

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