Question

The position x of a particle at time t is given by $x=\frac{v_0}{a}(1-e^{-at})$ where $v_0$ is a constant and a > 0. The dimensional formula of $v_0$ and a are

• A. $\left[M^0{L}^1{T}^{-1}\right]and\left[M^0{L}^0{T}^{-1}\right]$
• B. $\left[M^0{L}^1{T}^0\right]and\left[M^0{L}^0{T}^{-1}\right]$
• C. $\left[M^{0}L{T}^{-1}\right]and\left[M^0{L}^1{T}^{-2}\right]$
• D. $\left[M^0{L}^1{T}^{-1}\right]and\left[M^0{L}^0{T}^1\right]$

Answer 1

Answer: (A)

For finding $v_0=\frac{ax}{1-e^{a/t}}=\frac{m/s\times m/s^2}{1-e^{-m/s^2\times s}}=\frac{m^2}{s^2}\times \frac{s}{m}=m/s\left[M^{0}L{T}^{-1}\right]$

For finding,

$v_0=\frac{xa}{1-e^{-at}}\frac{v_0}{x}=\frac{a}{1-e^{-at}}\frac{x}{v_0}=\frac{1-e^{-at}}{a}a=\frac{v_0}{x}=\frac{m/s}{m}a=\frac{1}{s}a=s^{-1}=\left[M^0{L}^0{T}^{-1}\right]$