Question

The position of a particle at time $t$ is given by the relation $x=\frac{v_0}{\alpha }(1-e^{-\alpha t})$. The dimensional formula for ${\alpha }^2{v}_0^3$ will be :

Answer 1

$x=\frac{v_0}{\alpha }(1-e^{-\alpha t})....\left(A\right)$

From equation (A) since exponential is dimensionless

$\alpha t$ is dimensionless quantity

$\Rightarrow \alpha t=\left[M^0{L}^0{T}^0\right]$

$\therefore \left[\alpha \right]=\frac{\left[M^0{L}^0{T}^0\right]}{\left[T\right]}$

$\Rightarrow \left[\alpha \right]=\left[M^0{L}^0{T}^{-1}\right]...\left(1\right)$

Also,

$\left[\frac{v_0}{\alpha }\right]=\left[x\right]$

$\left[v_0\right]=\left[x\right]\left[\alpha \right]...\left(2\right)$

from (2) and (1)

$\left[v_0\right]=\left[x\right]\left[\alpha \right]=\left[L\right]\left[M^0{L}^0{T}^{-1}\right]$

$=\left[M^0{L}^1{T}^{-1}\right]....\left(3\right)$

Dimensional formula of ${\alpha }^2{v_0}^3$ can be obtained from from (1) and (3)

= $[M^0{L}^0{T}^{-1}{]}^2$ ${\left[M^0{L}^1{T}^{-1}\right]}^3$

$=\left[M^0{L}^3{T}^{-5}\right]$