Acceleration of a particle moving along the x-axis is defined by the law `a=-4x`, where a is in `m//s^(2)` and x is in meters. 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 its position coordinates.
(b) find its position x as function of time t.
(c) Find the maximum distance it can go away from the origin.

Options

A. `v=2sqrt(2-x^(2))`
B. `v=3sqrt(1-x^(2))`
C. `v=2sqrt(1-x^(3))`
D. `v=2sqrt(1-x^(2))`

Answer

D

Explanation

(a) By substituting given expression in the equation `a=v dv//dx` and rearranging, we have
`vdv=-4xdxrArr underset(2)overset(v)(int)vdv=-4 underset(0)overset(x)(int)xdx rArr v= +-2sqrt(1-x^(2)) rarr v=2sqrt(1-x^(2))`
since the particle passes the origin with positive velocity of `2 m//s`, so the minus sign in the eq. (i) has been dropped.
(b) By substituting above obtained expression of velocity in the equation `v=dx//dt` and rearranging, we have
`(dx)/sqrt(1-x^(2))=2dtrArr underset(0)overset(x)(int)(dx)/sqrt(1-x^(2))=2underset(0)overset(t)(int)dtrArr sin^(-1)(x)=2t rarr x=sin 2t`
(c) The maximum distance it can go away from the origin is `1m` because maximum magnitude of fine function is unity.