InterviewSolution
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InterviewSolution
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Two particles A and B start from rest at the origin `x=0` and move along a straight line such that `a_(A)=(6t-3)ms^(-2)` and `a_(B)=(12t^(2)-8)ms^(-2)` where t is in seconds based on the above facts answer the following questions, Total distance travelled by A at `t=4` s isA. 40 mB. 41 mC. 42 mD. 43 m |
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Answer» Correct Answer - B For A `dv_(A)=a_(A)dtimpliesint_(0)^(v_(A))dv_(A)=int_(0)^(t)(6t-3)dtimpliesV_(A)=3t^(2)-3t` `dv_(B)=a_(B)dtimpliesint_(0)^(v_(0))dv_(B)=int_(0)^(t)(12t^(2)-B)dt` `impliesV_(B)=4t^(3)-8t` Let us now calculate the times when A and B are at rest. The particle A is rest `(V_(A)=0)` when `3t^(2)-3t=0impliest=0` and `t=1`s The particle B is at rest `(V_(B)=0)`, when `4t^(3)-8t=0impliest=0s` and `t=sqrt(2)s` The position of particles A and B can be determined using `v=(dx)/(dt)` so `dx_(A)=V_(A)dt` `impliesint_(0)^(X_(A))dx_(A)=int_(0)^(t)(3t^(2)-3t)dt` `impliesX_(A)=t^(3)-(3)/(2)t^(2)` similarly `dX_(B)=V_(B)dt` `impliesint_(0)^(X_(B))dx_(B)=int-(0)^(t)(4t^(3)-8t)dtimpliesX_(B)=t^(4)-4t^(2)` The position of particle A at `t=1` s and 4 s are `X_(A)|t=1s``=1^(3)-(3)/(2)-(3)/(2)(1^(2))=0.5m` `X_(A)|t=4s``=4^(3)-(3)/(2)(4^(2))=40m` Particle A has travelled a total distance given by `d_(A)=2(0.5)+40=41m` The position of particle B at `t=sqrt(2)`s and 4 s are `X_(B)|t=sqrt(2)=(sqrt(2))^(4)-4(sqrt(2))^(2)=-4m` `X_(B)|t=4=(4)^(4)-4(4)^(2)=192m` Particle (B) has travelled a total distance given by `d_(B)=2(4)+192=200m` At `t=4s` s the distance between A and B is `DeltaX_(AB)=192-40=152m` |
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