Recovering velocity from an acceleration graph

Recovering velocity from an acceleration graph

6 marks
TGM Original
A particle has acceleration 2 m,s-2 for 0

Solution:

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Answer:\(v=2t-2\) for \(0\le t\le2\), then \(v=4-t\) for \(2\le t\le6\), in \(\mathrm{m\,s^{-1}}\). End velocities are \(-2,-2\); peak is \(2\) at \(2\) s. Reversals occur at \(1,4\) s.

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