Q. The displacement $x$ of a particle varies with time $t,\ x = ae^{- \alpha t} + be^{\beta t}$, 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
Explanation
(d) $x = ae^{- \alpha t} + be^{\beta t}$
Velocity $v = \frac{dx}{dt} = \frac{d}{dt}(ae^{- \alpha t} + be^{\beta t})$
$= a.e^{- \alpha t}( - \alpha) + be^{\beta t}(\beta) = - a\alpha e^{- \alpha t} + b\beta e^{\beta t}$
Acceleration $= - a\alpha e^{- \alpha t}( - \alpha) + b\beta e^{bt}.\beta$
$= a\ \alpha^{2}e^{- \alpha t} + b\beta^{2}e^{\beta t}$
Acceleration is positive so velocity goes on increasing with time
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.