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

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(D) In the equation $x(t) = \frac{v_0}{A}(1 - e^{-At})$, the exponent of the exponential function must be dimensionless. Therefore, the dimensions of

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$[M^0 LT^{-1}]$ and $[M^0 L^0 T^{-1}]$

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(D) In the equation $x(t) = \frac{v_0}{A}(1 - e^{-At})$, the exponent of the exponential function must be dimensionless. Therefore, the dimensions of $At$ must be $[M^0 L^0 T^0]$.Since $[t] = [T]$, we have $[A][T] = [1]$, which implies $[A] = [T^{-1}] = [M^0 L^0 T^{-1}]$.Next, the term $(1 - e^{-At})$ is dimensionless. Thus, the dimensions of $x$ must be equal to the dimensions of $\frac{v_0}{A}$.$[x] = \frac{[v_0]}{[A]} \implies [L] = \frac{[v_0]}{[T^{-1}]}$.Therefore, $[v_0] = [L][T^{-1}] = [M^0 LT^{-1}]$.Thus, the dimensions of $v_0$ and $A$ are $[M^0 LT^{-1}]$ and $[M^0 L^0 T^{-1}]$ respectively.

Column-$I$	Column-$II$
$(A)$ Electrical resistance	$(p)$ $M L^3 T^{-3} A^{-2}$
$(B)$ Electrical potential	$(q)$ $M L^2 T^{-3} A^{-2}$
$(C)$ Specific resistance	$(r)$ $M L^2 T^{-3} A^{-1}$
$(D)$ Specific conductance	$(s)$ None of these

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

Similar Questions

Dimensions of the following three quantities are the same:

Match the following two columns:

Column-$I$ Column-$II$

$(A)$ Electrical resistance $(p)$ $M L^3 T^{-3} A^{-2}$

$(B)$ Electrical potential $(q)$ $M L^2 T^{-3} A^{-2}$

$(C)$ Specific resistance $(r)$ $M L^2 T^{-3} A^{-1}$

$(D)$ Specific conductance $(s)$ None of these

The dimensional formula of $\frac{B^2}{2\mu_0}$ is . . . . . . .

The physical quantity that has the same dimensional formula as pressure is:

$[L^2 M^1 T^{-2}]$ are the dimensions of

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