A) \[-2av\]
B) \[-2av\]
C) \[-2av\]
D) None of these
Correct Answer: C
Solution :
Given that \[t=a{{s}^{2}}+bs+c\] On differentiating w.r.t. t, we get \[1=2as\frac{ds}{dt}+b\frac{ds}{dt}\] \[\Rightarrow \] \[1=2asv+bv\] ?(i) Again differentiating w.r.t. t, we get \[0=2a\frac{ds}{dt}v+2as\frac{dv}{dt}+b\frac{dv}{dt}\] \[\Rightarrow \] \[\frac{dv}{dt}(2as+b)=-2a{{v}^{2}}\] \[\Rightarrow \] \[\frac{dv}{dt}\left( \frac{1}{v} \right)=-2a{{v}^{2}}\] [from Eq. (i)] \[\Rightarrow \] \[\frac{dv}{dt}=-2a{{v}^{3}}\]
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