\(F = ma \)
\(10 = 2.5a \implies a = 4 ms^{-2}\)
Since it is a retarding movement, then \(a = -4 ms^{-2}\).
\(v = u + at; v = 0 ms^{-1}, u = 12 ms^{-1}\)
\(0 = 12 + (-4t) \implies 0 = 12 - 4t\)
\(4t = 12 \implies t = 3 s\)