InterviewSolution
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InterviewSolution
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Velocity of a particle in x-y plane at any time t is `v=(2t hati+3t^2 hatj) m//s` At `t=0,` particle starts from the co-ordinates `(2m,4m).` Find (a) acceleration of the particle at `t=1s.` (b) position vector and co-ordinates of the particle at `t=2s.` |
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Answer» Correct Answer - A::B::C::D (a) `a=(dv)/dt=d/dt(2t hati+3t^2hatj)` `=(2hati++6thatj) m//s^2` At `t=1s,` `a=(2hati+6 hatj) m//s^2` (b) intds=intv dt or `s=intv dt=int(2 thati+3t^2 hatj)dt` `:. r_f-r_i=int_(i ntitial)^(fi nal) (2t hati+3t^2 hatj)dt` `or r_(2 sec)-r_(0 sec)=int_0^2(2t hati+3t^2 hatj)dt` `:. r_(2 sec)=r_(0 sec)+[t^2 hati+t^3 hatj]_0^2` `=(2 hati+4 hatj)+(4 hati+8 hatj)` `=(6 hati+12 hatj) m` Therefore, coordinates of the particle at `t=2s` are `(6m,12m)` |
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