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Acceleration of particle moving along th...

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?

Text Solution

Verified by Experts

(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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