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(D) In the equation $x(t) = \frac{v_0}{A}(1 - e^{-At})$,the exponent of the exponential function must be dimensionless. Therefore,the dimension of $At

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$[M^0 LT^{-1}]$ and $[M^0 L^0 T^{-1}]$

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(D) In the equation $x(t) = \frac{v_0}{A}(1 - e^{-At})$,the exponent of the exponential function must be dimensionless. Therefore,the dimension of $At$ must be $1$. $[A][T] = [M^0 L^0 T^0] \implies [A] = [T^{-1}] = [M^0 L^0 T^{-1}]$. Since the term $(1 - e^{-At})$ is dimensionless,the dimension of $x$ must be equal to the dimension of $\frac{v_0}{A}$. $[x] = \frac{[v_0]}{[A]} \implies [L] = \frac{[v_0]}{[T^{-1}]} \implies [v_0] = [L][T^{-1}] = [M^0 LT^{-1}]$. Thus,the dimensions of $v_0$ are $[M^0 LT^{-1}]$ and the dimensions of $A$ are $[M^0 L^0 T^{-1}]$.

The position of a particle at time $t$ is given by the equation $x(t) = \frac{v_0}{A}(1 - e^{-At})$,where $v_0$ is a constant and $A > 0$. The dimensions of $v_0$ and $A$ respectively are:

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