A) \[{{t}^{-1}}\]
B) \[{{t}^{-1/2}}\]
C) \[{{t}^{1/2}}\]
D) \[t\]
Correct Answer: B
Solution :
Here, \[KE=\frac{1}{2}m{{v}^{2}}=kt\] where, k is a constant Differentiating w.r.t., t we get \[\frac{1}{2}m\,2v\,\frac{dv}{dt}=k\] Or \[m\,\frac{dv}{dt}=\frac{k}{v}\] Or \[F=\frac{k}{v}\] But \[v=\sqrt{\frac{2kt}{m}}\] \[\therefore \] \[F=\frac{k}{\sqrt{\frac{2kt}{m}}}\] Or \[F\propto {{t}^{-1/2}}\]You need to login to perform this action.
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