The displacement x of a particle varies with time according to the relation `x=(a)/(b)(1-e^(-bt))`. Then select the false alternative.

Options

A. ` At t=1 //b` the displacement of the particle is nearly (2/3) (a/b)
B. The velocity and acceleration of the particle at t = 0 are respectively
C. The particle cannot reach a point at a distance x' from its starting position if x' gt a/b
D. The particle will come back to its starting point as` t to infinity `

Answer

A, B, C

Explanation

Velocity of the particle is given by `v =(dx) /(dt) =(d)/(dt) {(a)/(b) (1-e^(-bt) ) }=ae^(-bt) `
Acceleration of the particle is given by ` alpha =(dv)/(dt) =(d)/(dt) (ae^(-bt) ) =-abe^(-bt) `
At t = `1//b` the displacement of the particle is ` x= ( a)/(b) (1-e^(-1) ) =(a)/(b) (1-(1)/(3)) =(2a)/(3b) (because e^(-1) =(1)/(3))`
Hence choice (A) is correct. At t = 0, the value v and ` alpha ` respectively are ` v=ae^(-0) = a and alpha =abe^(-0) . `Hence choice (B) is also correct. The displacement x ismaximum when ` t to infinity , i.e. x _(max) =(a)/(b) (1-e^(-infinity))=(a)/(b) .Hence choice (C) is also correct . Thus the correct choice are (a) , (b) and (C)