Question

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

• A. $M^{0} L^{1} T^{-1}$ and $T^{-1}$
• B. $M^{0} L^{1} T^{0}$ and $T^{-1}$
• C. $M^{0} L^{1} T^{-1}$ and $L T^{-2}$
• D. $M^{0} L^{1} T^{-1}$ and $T$

Answer 1

Answer: (A)

Dimension of $\alpha t=\left[M^{0} L^{0} T^{0}\right] $
$\therefore[\alpha]=\left[T^{-1}\right]$
Again $\left[\frac{v_{0}}{\alpha}\right]=[L]$
so $\left[v_{0}\right]=\left[L T^{-1}\right]$