Maximum displacement and intervals of negative acceleration from a velocity function by technology
Maximum displacement and intervals of negative acceleration from a velocity function by technology
7 marks
A particle travels in a straight line, starting from a fixed point \(O\) on the line. Its velocity \(t\) seconds after it starts is \(v\text{ m s}^{-1}\), where
\(v=\left(2t^2-9t+6\right)\mathrm{e}^{-0.6t}+2\sin t,\quad 0\leq t\leq8.\)
- Find the greatest displacement of the particle from \(O\) during the motion.[4]
- Find the set of times for which the particle has negative acceleration.[3]
Solution:
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Answer:(a) \(4.28\text{ m}\) (at \(t=8\)) (b) \(0\leq t<4.54\) and \(7.80<t\leq8\)
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