InterviewSolution
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InterviewSolution
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The motion of a body is given by the equation `dv//dt=6-3v`, where v is in m//s. If the body was at rest at `t=0` (i) the terminal speed is `2 m//s` (ii) the magnitude of the initial acceleration is `6 m//s^(2)` (iii) The speed varies with time as `v=2(1-e^(-3t)) m//s` (iv) The speed is `1 m//s`, when the acceleration is half initial value |
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Answer» (a) `(dv)/(dt)=6-3v=3(2-v)` `int_(0)^(v)(dv)/((2-v))=3int_(0)^(t) dt` `(| log_(e)(2-v)|_(0)^(v))/-1=3|t|_(0)^(t)` `log_(e)(2-v)-log_(e)v=-3t` `log_(e)((2-v)/v)=-3t` `(2-v)/v=e^(-3t)` `2-v=2e^(-3t)` `v=2(1-e^(-3t))` (b) Terminal velocity, i.e. at `t=oo`, `v=2(1-e^(-3t))=2(1-e^(-oo))` , `=2(1-1/e^(oo))=2(1-1/oo)` `=2(1-0)=2m//s` |
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