a=dv/dt=4t-1.
Integrating:
v(t)=2t²-t+c₁, v(0)=c₁=5m/s, v(t)=2t²-t+5.
v=ds/dt=2t²-t+5.
Integrating:
s(t)=2t³/3-t²/2+5t+c₂, s(0)=c₂=2m, s(t)=2t³/3-t²/2+5t+2.
s(7)=686/3-49/2+35+2=1447/6 m.
Distance travelled from t=0 (when s=2m) to t=7=1435/6 m.
This can also be calculated from the definite integral:
∫₀⁷(2t²-t+5)dt=[2t³/3-t²/2+5t]₀⁷=(686/3-49/2+35)=1435/6 m.
(Note that at t=0, a=-1m/s², v=5m/s, s=2m; and at t=7, a=27m/s², v=96m/s, s=1447/6 m.)