The position of a particle at time t is given | vedclass

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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 dimensions of

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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 dimensions of $At$ must be $[M^0 L^0 T^0]$.Since $[t] = [T]$, we have $[A][T] = [1]$, which implies $[A] = [T^{-1}] = [M^0 L^0 T^{-1}]$.Next, the term $(1 - e^{-At})$ is dimensionless. Thus, the dimensions of $x$ must be equal to the dimensions of $\frac{v_0}{A}$.$[x] = \frac{[v_0]}{[A]} \implies [L] = \frac{[v_0]}{[T^{-1}]}$.Therefore, $[v_0] = [L][T^{-1}] = [M^0 LT^{-1}]$.Thus, the dimensions of $v_0$ and $A$ are $[M^0 LT^{-1}]$ and $[M^0 L^0 T^{-1}]$ respectively.

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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