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The displacement $x$ of a particle varies with time $t,x = a{e^{ - \alpha t}} + b{e^{\beta t}}$, where $a,\,b,\,\alpha \,{\rm{and }}\beta $ are positive constants. The velocity of the particle will
Go on decreasing with time
Be independent of $\alpha $ and $\beta $
Drop to zero when $\alpha = \beta $
Go on increasing with time
Solution
(d) $x = a{e^{ – \alpha t}} + b{e^{\beta t}}$
Velocity $v = \frac{{dx}}{{dt}} = \frac{d}{{dt}}(a{e^{ – \alpha t}} + b{e^{\beta t}})$
$ = a.{e^{ – \alpha t}}( – \alpha ) + b{e^{\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.
