Acceleration of particle moving rectilinearly is `a = 4-2x`(where x is position in metre and a in ms^(-2))`. It si at instantaneous rest at x = 0. At what position `x 9`in meter) will the particle again come to instantaneous rest?

Answer

4

Explanation

`(vdv)/(dx) = 4-2x`
`int_(0)^(v) vdv = int_(0)^(x) (4-2x)dx rArr v^2/2 = 4x - x^2`
when `v = 0 , 4x - x^2 = 0`
`x = 0, 4`
`rArr Thus, at x = 4, the particle will again come to rest.