Question

The displacement $x$ of a particle varies with time $t$, $x=a{e}^{-pt}+b{e}^{qt}$, where $a,b,p$ and $q$ are positive constants. The velocity of the particle will

• A. go on increasing with time
• B. be independent of $p$ and $q$
• C. drop to zero when $p=q$
• D. go on decreasing with time

Answer 1

Answer: (A)

Given, $x=a{e}^{-pt}+b{e}^{qt}$
Velocity, $v=\frac{dx}{dt}=\frac{d}{dt}\left(a{e}^{-pt}+b{e}^{qt}\right)$
$=-pa{e}^{-pt}+qb{e}^{qt}$
Acceleration, $a=\frac{dv}{dt}=\frac{d}{dt}\left(-pa{e}^{-pt}+qb{e}^{qt}\right)$
$=p^{2}a{e}^{-pt}+q^{2}b{e}^{qt}$
Acceleration is positive, so velocity goes on increasing with time.