The position of x of a particle at time t is given by x - V 0 - a 1 - e - a t where V 0 is constant and a-0 The dimensions of V 0 and a are

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Dimensional Analysis

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Question

The position of x of a particle at time t is given by $x = \frac{V_{0}}{a} (1 - e^{- a t})$ where $V_{0}$ is constant and a>0 The dimensions of $V_{0}$ and a are

M^{0} L T^{- 1} and T^{- 1}

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M^{0} L T^{- 0} and T^{- 1}

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M^{0} L T^{- 1} and L T^{- 2}

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M^{0} L T^{- 1} and T

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Solution

The correct option is A $M^{0} L T^{- 1} and T^{- 1}$
The position of the particle at a time t is given as

$X (t) = \frac{V_{0}}{a (1 - e^{- a t})}$
As X denotes position
$[X] = m$ here, $[a] = s^{- 1} \to a = [M^{0} L^{0} T^{- 1}]$

$\Rightarrow \frac{[V_{0}]}{[a]} = [V_{0}] s = m$

Hence, $V_{0}$ has dimensions of velocity
$[V_{0}] = m s^{- 1} V_{0} = [M^{0} L^{1} T^{- 1}]$
so the correct option is $A .$

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The position of x of a particle at time t is given by x=V0a(1−e−at) where V0 is constant and a>0 The dimensions of V0 and a are

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The position of x of a particle at time t is given by $x = \frac{V_{0}}{a} (1 - e^{- a t})$ where $V_{0}$ is constant and a>0 The dimensions of $V_{0}$ and a are