The acceleration of a particle as it moves along a straight line is given by a = (4t – 1) m/image, where t is in seconds. If s = 2 m and v = 5 m/s when t = 0, determine the particle’s velocity and position when t = 7s. Also, determine the total distance the particle travels during this time period.

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v(t)=2t²-t+c₁, v(0)=c₁=5m/s, v(t)=2t²-t+5.



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.)

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