Question

The position of a particle at time t is given by the relation $x\left(t\right)=\left(\frac{v_0}{\alpha }\right)(1-c^{-\alpha t})$, where $v_0$ is a constant and $\alpha >0$. Find the dimensions of $v_0$ and $\alpha$.

Answer 1

$x\left(t\right)=\left(\frac{V_0}{\alpha }\right)(1-C^{-\alpha t})$

$V_0$ is constant and $\alpha >0$

Since, $C$ is constant so $\alpha t$ should be unitless

So unit $\alpha ={\sec }^{-1}=\left[T^{-1}\right]$

$x$ is length, $x$ is in metre.

$V_0=x\alpha$

$V_0=M{sec}^{-1}$

$V_0=\left[{LT}^{-1}\right]$

$\alpha =\left[T^{-1}\right]$