The displacement of a particle after time t is given by `x = (k // b^(2)) (1 - e^(-bi))`. Where b is a constant. What is the acceleration fo the particle ?

Options

A. `ke^(-bt)`
B. `-ke^(-bt)`
C. `(k)/(b^(2))e^(-bt)`
D. `(-k)/(b^(2))e^(-bt)`

Answer

B

Explanation

Velocity at the time of entering = `sqrt(2gh)`
`F_("net")=d_(2)vg-d_(1)vg`
`a=((d_(2)-d_(1))vg)/(d_(1)v)`
`implies a=((d_(2)-d_(1))/(d_(1)))g`
`implies v_(F)=sqrt(2gh)-at`
`O=sqrt(2gh)-((d_(2)-d_(1))/(d_(1)))g timplies t=(sqrt(2gh))/(g)(d_(1))/((d_(2)-d_(1)))=sqrt((2h)/(g))((d_(1))/(d_(2)-d_(1)))`