s=½at² where s is distance, a is acceleration, t=time.
s=120, a=2.6, so t²=2s/a=240/2.6.
At that time v, velocity=at=2.6√(240/2.6)=√(240×2.6)=√624 m/s.
The motorcycle now decelerates, so a=-1.5, v=u+at gives us the final velocity v for initial velocity u, so we have v=12, u=√624 and a=-1.5. We have the formula v²-u²=2as, so:
144-624=-3s, therefore s=(624-144)/3=480/3=160m.
While the motorcycle was decelerating it covered a distance of 160m and it had already travelled 120m before decelerating, so the total distance for the whole trip was 160+120=280m.