InterviewSolution
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InterviewSolution
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The speed of a particle moving in a circle of radius `r=2m` varies witht time `t` as `v=t^(2)`, where `t` is in second and `v` in `m//s`. Find the radial, tangential and net acceleration at `t=2s`. |
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Answer» Given in the question, `v=t^(2)` `therefore` Linear speed of particle at `t=2s` is `v=(2)^(2)=4ms^(-1)` `therefore" Radial acceleration "a_(r)=(v^(2))/(r)=((4)^(2))/(2)=8ms^(-2)` The tangential acceleration is `a_(t)=(dv)/(dt)=(d)/(dt)(t^(2))=2t` `therefore` Tangential acceleration at t 2 s is `a_(t)=(2)(2)=4ms^(-2)` `therefore` Net acceleration of particle at t = 2 s is `a=sqrt((a_(r))^(2)+(a_(t))^(2))=sqrt((8)^(2)+(4)^(2))=sqrt(64+16)" or "a=sqrt(80)"ms"^(-2)` |
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