InterviewSolution
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InterviewSolution
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Acceleration of particle moving along the x-axis varies according to the law `a=-2v`, where a is in `m//s^(2)` and v is in `m//s`. At the instant `t=0`, the particle passes the origin with a velocity of `2 m//s` moving in the positive x-direction. (a) Find its velocity v as function of time t. (b) Find its position x as function of time t. (c) Find its velocity v as function of its position coordinates. (d) find the maximum distance it can go away from the origin. (e) Will it reach the above-mentioned maximum distance? |
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Answer» (a) By substituting the given relation `a=dv//dt`, we have `(dv)/v=-2dtrArr underset(2)overset(v)(int)(dv)/v=-2underset(0)overset(t)(int)dt rarr v=2e^(-2t)`…(i) (b) By substituting the above equation in `v=dx//dt`, we have `dx=2e^(-2t)dt rArr underset(0)overset(x)(int)dx=2underset(0)overset(t)(int)e^(-2t)dt rarr x=1-e^(-2t)`...(ii) (c) Substituting given expression a in the equation `a=vdv//dx` and rearranging, we have `dv=-2dxrArr underset(2)overset(v)(int)dv=-2underset(0)overset(x)(int)dx rarr v=2(1-x)` ...(iii) (d) Eq. (iii) suggest that to cover `1m` it will take whose value tends to infinity. Therefore, it can never cover this distance. |
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