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The position of a particle at time ‘t’ is given by the relation $x (t) = (\frac{V_{0}}{α}) (1 - e^{- α t})$ , where $V_{0}$ is a constant and $α > 0.$ Then determine the dimensions of $V_{0} a n d α .$

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$[M^{0} L^{1} T^{0}] a n d [T^{- 1}]$

$[M^{0} L^{1} T^{- 1}] a n d [T^{- 1}]$

$[M^{0} L^{1} T^{- 1}] a n d [T^{- 2}]$

$[M^{0} L^{1} T^{- 1}] a n d [T]$

answer is B.

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Detailed Solution

$[α] = {[T]}^{- 1} [\frac{V_{o}}{α}] = [L] [V_{o}] = [L] {[T]}^{- 1}$

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