Question

The displacement x of a particle varies with time t as $x=ae^{- \alpha t} + be ^{\beta t}$, where $a,b,$ $\alpha$ and $\beta$ are positive constants. The velocity of the particle will:

• 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 1

Answer: (D)

go on increasing with time

Answer 2

Given x=$$\,ae^{-\alpha t}+be^{\beta t}

Where $a,b$,$\,\alpha$, and $\beta$ are positive constant

$V=$ $\frac{dx}{dt}$ =$\frac{d(ae^{-\alpha t}+be^{\beta t})}{dt}$ =$−aαe^{ −αt}+ bβe^{βt }$

∴ $\frac{dx}{dt}$=$aα^2e^{-αt}+bβ^2e^{\beta t} \text{ is always >0}$

V is increasing the function of t.

Therefore, the correct option is (D): go on increasing with time.