The position of a particle at time t is given by the relation x(t)=(v₀/a)(1-c ⁻ᵃᵗ) where v₀ is a constant and α > 0 . The dimensions of v₀ and a are respectively

The correct option is M⁰L¹T⁻¹ and T⁻¹.

To find the dimensions of $v_0$ and $a$, we analyze the given relation for the position of a particle at time $t$:

$x\left(t\right)=\frac{v_0}{a}(1-e^{-at})$

Note: The term $c^{-at}$ in the question is equivalent to the exponential function $e^{-at}$ (where $a$ represents the constant $\alpha$).

Step 1: Determine the dimensions of $a$

The power of an exponential function must be a dimensionless quantity. Therefore, the term $at$ is dimensionless:

$\left[at\right]=\left[M^0{L}^0{T}^0\right]$

Using the dimension of time $\left[t\right]=T$:
$\left[a\right]·T=1$
$\left[a\right]=T^{-1}$

So, the dimensional formula for $a$ is $M^0{L}^0{T}^{-1}$.

Step 2: Determine the dimensions of $v_0$

Since the term $(1-e^{-at})$ is dimensionless, the dimension of position $x\left(t\right)$ must equal the dimension of the ratio $\frac{v_0}{a}$:

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

Since position $x$ has the dimension of length ($L$):
$L=\frac{\left[v_0\right]}{T^{-1}}$
$\left[v_0\right]=L·T^{-1}=M^0{L}^1{T}^{-1}$

Therefore, the dimensions of $v_0$ and $a$ are $M^0{L}^1{T}^{-1}$ and $T^{-1}$ respectively.