The displacement `x` of a particle varies with time `t` as `x = ae^(-alpha t) + be^(beta t)`. Where `a,b, alpha` and `beta` positive constant.
The velocity of the particle will.

Options

A. go on decreasing with time
B. be independent of `alpha` and `beta`
C. drop to zero when `alpha=beta`
D. go on increasing with time

Answer

D

Explanation

Given, `x=ae^(-alpha t)+be^(beta t)`
So, velocity, `upsilon=(dx)/(dt)=-a alpha e^(-alpha t)+b beta e^(beta t)`
`a=(d upsilon)/(dt)=a alpha^(2)e^(-alpha t)+b beta^(2)e^(beta t)=ve` all time.
`therefore upsilon` will go on increasing.