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

Question

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

Aicharya · Accepted Answer

Here, $a t$ is dimensionless $\Rightarrow a=\frac{1}{t}=\left[\frac{1}{T}\right]=\left[T^{-1}\right]$ $x=\frac{V_{0}}{a}$ and $V_{0}=x a=\left[L T^{-1}\right]=\left[M^{0} L T^{-1}\right]$ ^

The position $x$ of a particle at time $t$ is given by $x=\frac{V_{0}}{a}\left(1-e^{-\alpha t}\right)$,…

Solution