The position of a particle at time t is given | vedclass

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(A) The argument of an exponential function must be dimensionless. Therefore,the dimension of $\alpha t$ must be $[M^0 L^0 T^0]$.Since $[t] = [T]$,we

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$M^0 L^1 T^{-1}$ and $T^{-1}$

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(A) The argument of an exponential function must be dimensionless. Therefore,the dimension of $\alpha t$ must be $[M^0 L^0 T^0]$.Since $[t] = [T]$,we have $[\alpha] [T] = [1]$,which gives $[\alpha] = [T^{-1}]$.In the expression $x(t) = (v_0 / \alpha) (1 - e^{-\alpha t})$,the term $(1 - e^{-\alpha t})$ is dimensionless.Therefore,the dimension of $x(t)$ must be equal to the dimension of $(v_0 / \alpha)$.$[x] = [L]$,so $[v_0 / \alpha] = [L]$.$[v_0] = [L] 	imes [\alpha] = [L] 	imes [T^{-1}] = [L T^{-1}]$.Thus,the dimensions of $v_0$ and $\alpha$ are $[M^0 L^1 T^{-1}]$ and $[T^{-1}]$ respectively.

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

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