A) \[{{t}^{1/2}}\]
B) \[{{t}^{3/4}}\]
C) \[{{t}^{3/2}}\]
D) \[{{t}^{2}}\]
Correct Answer: C
Solution :
Power \[P=\frac{dW}{dt}=\frac{F.ds}{dt}\] ( \[\therefore \] \[dW=F.ds\]) \[=m\frac{dv}{dt}.\frac{ds}{dt}\] \[=\frac{mvdv}{dt}\] \[\therefore \] P is constant, say K then \[\frac{mvdv}{dt}=K\] or \[mvdv=Kdt\] Integrating, we get \[=\frac{m{{v}^{2}}}{2}=Kt+c\] Using boundary condition \[c=o\] So, \[\frac{m{{v}^{2}}}{2}=Kt\] or \[v=\sqrt{2\frac{Kt}{m}}\] or \[\frac{ds}{dt}=\sqrt{\frac{2Kt}{m}}\] or \[ds=\sqrt{\frac{2K}{m}}.{{t}^{1/2}}dt\] On integrating, we get \[s=\sqrt{\frac{2K}{m}}\frac{2}{3}{{t}^{3/2}}\] \[\Rightarrow \] \[s\propto {{t}^{3/2}}\]
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