Example 4: Dopey Dan starts from rest at a traffic light and accelerates at 3.6 m/s2. 1.2 seconds later Speedy Sue passes through the same light moving at 2.7 m/s and accelerating at 4.2 m/s2. Where does she catch Dan?

Solution: I'll take the origin at the light and positive forward...
 

Dan Sue
a = 3.6 m/s2
vo = 0
so = 0
tDan = t
a = 4.2 m/s2
vo = 2.7 m/s
so = 0
tSue = t - 1.2

Subtracting these equations, I get

Solving this using the quadratic formula,

I discard the negative time, since it has no physical meaning. Substituting the 7.89 (unrounded) into Dan's equation (the easier equation), I get 112 m. So Sue catches Dan 112 m from the light.