Position of a particle moving along straight line is given by `x(t)=(A)/(B)(1-e^(-At))`, where `B` is constant and `Agt0`. Dimension of `(A^(3))/(B)` is similar to `:-`

Options

A. (a)Linear momentum
B. (b)Moment of inertia
C. (c)Relative velocity
D. (d)Acceleration

Answer

D

Explanation

At is dimensionless so `[At]=1`
`implies[A]=[T^(-1)]` also `[x]=[(A)/(B)]` so `[B]=[L^(-1)T^(-1)]` Therefore `[A^(3)/(B)]=[(T^(-2))/(L^(-1)T^(-1)]]=[LT^(-2)]`