The position of a particle at time $t$ is given by the equation $x(t)=\frac{v_0}{A}\left(1-e^{A t}\right)…
The position of a particle at time $t$ is given by the equation $x(t)=\frac{v_0}{A}\left(1-e^{A t}\right) v_o=$ constant and $A>0$. Dimensions of $v_o$ and A respectively are
$\left[\mathrm{M}^0 \mathrm{LT}^0\right]$ and $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]$
$\left[\mathrm{M}^0 \mathrm{LT}^{-1}\right]$ and $\left[\mathrm{M}^0 \mathrm{LT}^{-2}\right]$
$\left[\mathrm{M}^0 \mathrm{LT}^{-1}\right]$ and $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}\right]$
$\left[\mathrm{M}^0 \mathrm{LT}^{-1}\right]$ and $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]$
Solution
\(x(t)=\left(\frac{V_0}{\alpha}\right)\left(1-C^{-\alpha t}\right)\)
\(\mathrm{V}_{0}\) is constant and \(\alpha > 0\)
Since, \(e\) is constant so \(\alpha\) t should be unitless
So unit \(\alpha=\sec ^{-1}=\left[\mathrm{T}^{-1}\right]\) x is length, x is in metre.
$\begin{aligned}
& V_{0}=x \alpha \\
& V_{0}=M\sec^{-1} \\
& V_{0}=\left[LT^{-1}\right] \\
& \alpha=\left[T^{-1}\right]
\end{aligned}$
Asked in: JEE Mains - Units and Dimensions - Test 2