If a,b and c are constant of motion and `s = at^(2)+bt+ c`, then prove that 4a(s-c) = `v^(2) - b^(2)`.

Explanation

Here , s = `at^(2)` +bt +c
`:. " " v (ds)/(dt) `= 2at + b
`:. " " v^(2)-b^(2) = (2at+b)^(2)-b^(2) = 4a^(2)t^(2) +4abt`
`= 4a(at^(2)+bt) = 4a(at^(2)+bt+c-c)`
or, `" " v^(2)-b^(2) = 4a(s-c)` (Proved).