4 A point P on a piece of machinery is moving in a vertical straight line. The displacement of P above ground level at time \(t\) seconds is \(y\) metres. The displacement-time graph for the motion during the time interval \(0 \leqslant t \leqslant 4\) is shown in Fig. 7 .
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\includegraphics[alt={},max width=\textwidth]{34e4ce80-21b0-48f5-865c-de4dd837f7c5-3_1027_1333_372_435}
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\caption{Fig. 7}
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- Using the graph, determine for the time interval \(0 \leqslant t \leqslant 4\)
(A) the greatest displacement of P above its position when \(t = 0\),
(B) the greatest distance of P from its position when \(t = 0\),
(C) the time interval in which P is moving downwards,
(D) the times when P is instantaneously at rest.
The displacement of P in the time interval \(0 \leqslant t \leqslant 3\) is given by \(y = - 4 t ^ { 2 } + 8 t + 12\). - Use calculus to find expressions in terms of \(t\) for the velocity and for the acceleration of P in the interval \(0 \leqslant t \leqslant 3\).
- At what times does P have a speed of \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the interval \(0 \leqslant t \leqslant 3\) ?
In the time interval \(3 \leqslant t \leqslant 4 , \mathrm { P }\) has a constant acceleration of \(32 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). There is no sudden change in velocity when \(t = 3\).
- Find an expression in terms of \(t\) for the displacement of P in the interval \(3 \leqslant t \leqslant 4\).