The Position Vector of an Object of Mass 0

The position vector of an object of mass 0.50 kg subject to a constant force is given by = (at²+bt) + (ct²+dt) + (et²+ft) , where a = 2.0 m/s², b = 3.0 m/s, c = 2.5 m/s², d = -2.0 m/s, e = 1.0 m/s², and f = 4.0 m/s. What is the angular momentum of the object about the origin at t = 2.0 s?

A) (24 - 122
- 89
) kg∙m²/s
B) (-24 + 10
+ 23
) kg∙m²/s
C) (25 - 14
+ 20
) kg∙m²/s
D) (25 + 14
+ 20
) kg∙m²/s
E) (24 + 122
+ 23
) kg∙m²/s