A man has to go 50 m due north, 40 m due east and 20 m due south to reach a field,
(A) What distance he has to walk to reach the field?
(B)What is his displacement from his house to the field?

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.