A particle has velocity v=4-2t m,s-1 for 0 <= t 4 seconds and initial displacement 1 m. Sketch its displacement–time, distance–time, velocity–time, speed–time and acceleration–time graphs, labelling endpoint values and any rest time. Distance means total distance since t=0.
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Answer:\(s=1+4t-t^2\) m; distance \(D=4t-t^2\) before \(2\) s and \(D=t^2-4t+8\) afterwards. Velocity \(4-2t\), speed \(|4-2t|\) in \(\mathrm{m\,s^{-1}}\); acceleration \(-2\ \mathrm{m\,s^{-2}}\). Rest at \(2\) s; endpoint distances \(0,8\) m.