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條題目係
A particle moves in a straight line,starting with a velocity of 4m/s .After s seconds , its acceleration is (6t-8) m/ss  
佢問: the total time that elapses before the particle returns to its starting point? 同 the greatest velocity of the particle in the interval ?
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Well, if you can draw the V-t picture, it is better to do.
However, it's hard to draw in the condition.
Sol.
     Vo = 4 (m/s)
     a = (6t-8) (m/s^2)
     => V(t) = 4 + 3t^2  - 8t  = 3(t^2 - 8/3 t + 16/9) -4/3 = 3(t - 4/3)^2 - 4/3   
     => S(t) = 4t + t^3  - 4t^2  => S = 0 => t(4+t^2 -4t)=0 => t = 0, 2
     ----> t=2 (s), come back, and Vmax = 4  at t=0.