| Direction: A particle initially (i.e., at time \[t=0\]) moving with a velocity u subjected to a retarding force, as a result of which it decelerates at a rate \[a=-k\,\sqrt{v}\] where v is the instantaneous velocity and k is a positive constant. |
| The particle comes to rest in a time |
A) \[\frac{2\sqrt{u}}{k}\]
B) \[\frac{\sqrt{u}}{k}\]
C) \[2k\sqrt{u}\]
D) \[k\sqrt{u}\]
Correct Answer: A
Solution :
| [a] Given \[a=\,-\,k{{v}^{1/2}}or\,\frac{dv}{dt}=-k{{v}^{1/2}}\] | |
| Thus, \[{{v}^{-1/2}}dv=\,-kdt\] | |
| Integrating, we have | |
| \[\int{{{v}^{-1/2}}dv=-k\int_{{}}^{{}}{dt}}\] | |
| or \[2{{v}^{1/2}}=-kt+c\] | ...(i) |
| where c is the constant of integration, Given that at \[t=0,\text{ }v=0\]. Using this in Eq. (i), we get \[2{{u}^{1/2}}=c\]. Using this value of c in Eq. (i), we have | |
| \[2({{v}^{1/2}}{{u}^{1/2}})=kt\] | ...(ii) |
| Let \[\tau \] be the time taken by the particle to come to rest, | |
| Then, \[v=0\] at \[t=\tau \], | |
| Using this in Eq. (ii), we get | |
| \[2(0-{{u}^{1/2}})=-k\tau \,\,or\,\tau =\frac{2{{u}^{1/2}}}{k}\] | ...(iii) |
| Hence, the correct choice is [a]. | |
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