| 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 distance covered by the particle before coming to rest is |
A) \[\frac{{{u}^{3/2}}}{k}\]
B) \[\frac{2{{u}^{3/2}}}{k}\]
C) \[\frac{3{{u}^{3/2}}}{2k}\]
D) \[\frac{2{{u}^{3/2}}}{3k}\]
Correct Answer: D
Solution :
| [d] To find the distance s covered in this time, we use |
| Eq. (i) to get |
| \[{{v}^{1/2}}=\text{ }{{u}^{1/2}}\frac{kt}{2}\] |
| Squaring, we have |
| \[v=u-kt{{u}^{1/2}}\,+\frac{{{k}^{2}}{{t}^{2}}}{4}\] |
| But \[v=\frac{ds}{dt}\] |
| Therefore, \[\frac{ds}{dt}=u-kt\,\,{{u}^{1/2}}\,+\frac{{{k}^{2}}{{t}^{2}}}{4}\] |
| Integrating from |
| \[t=0\] to \[t=\tau \] we have |
| \[s=\left| ut-\frac{k{{u}^{1/2}}{{t}^{2}}}{2}+\frac{{{k}^{2}}{{t}^{2}}}{12} \right|_{0}^{\tau }\] |
| or \[s-\,u\tau -\frac{1}{2}\,k{{u}^{1/2}}{{\tau }^{2}}+\,\frac{1}{12}{{k}^{2}}{{\tau }^{3}}\] ... (iv) |
| Substituting the value of t from Eq. (iii) in Eq. (iv), we get |
| \[s=\,\frac{2{{u}^{3/2}}}{k}-\,\frac{4{{u}^{3/2}}}{2k}+\frac{8{{u}^{3/2}}}{12k}\] |
| or \[s=\,\frac{2{{u}^{3/2}}}{3k}\] |
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