Question

The displacement $x$ of a particle varies with time $t$, $x=a e^{-p t}+b e^{q t}$, 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, $x=a e^{-p t}+b e^{q t}$
Velocity, $v=\frac{d x}{d t}=\frac{d}{d t}\left(a e^{-p t}+b e^{q t}\right)$
$=-p a e^{-p t}+q b e^{q t}$
Acceleration, $a=\frac{d v}{d t}=\frac{d}{d t}\left(-p a e^{-p t}+q b e^{q t}\right)$
$=p^{2} a e^{-p t}+q^{2} b e^{q t}$
Acceleration is positive, so velocity goes on increasing with time.