A particle starts from point `A`, moves along a straight line path with an accleration given by `a = 2 (4 - x)` where `x` is distance from point `A`. The particle stops at point `B` for a moment. Find the distance `AB` (in `m`). (All values are in `S.I.` units)

Answer

8

Explanation

`v(dv)/(dx) = 8 - 2x`
`rArr overset(v)underset(0)(int)vdv = overset(x)underset(0)int(8 - 2x)dx rArr (v^(2))/(2) = 8x - x^(2)`
`rArr v^(2) = 16x - 2x^(2)`
At `B, v = 0` so, `x = 8`
Hence, `AB = 8`