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0606 P13 - Jun 2025 - Q11 - 10 marks
7137
A particle \(P\) moves in a straight line and passes through a fixed point \(O\). At time \(t\) seconds, its displacement from \(O\), \(s\) metres, is given by
\(s=t+6t^2-t^3\) for \(0\leqslant t\leqslant3\).
\(s=12t-\frac13t^2-3\) for \(3\leqslant t\leqslant k\), where \(k\) is a constant.
It is given that, for \(3\leqslant t\leqslant k\), the velocity of \(P\) is positive and its acceleration is negative.
(a) The maximum velocity of \(P\) occurs when \(t=2\). On the axes below, sketch a velocity-time graph for the first \(k\) seconds of the motion of \(P\).
(b) The total distance travelled by \(P\) for \(0\leqslant t\leqslant k\) is 57 metres. Given that when \(t=3\) the distance and displacement of \(P\) from \(O\) are equal, find the value of \(k\).
Solution
Answer: \(k=6\).
For \(0\leqslant t\leqslant3\),
\(s=t+6t^2-t^3.\)
So
\(v=\frac{\mathrm{d}s}{\mathrm{d}t}=1+12t-3t^2.\)
This is a downward-opening parabola. It has \(v(0)=1\) and \(v(3)=10\), and its maximum occurs at \(t=2\).
For \(3\leqslant t\leqslant k\),
\(s=12t-\frac13t^2-3.\)
So
\(v=12-\frac23t.\)
This is a straight line with negative gradient, starting from \(v(3)=10\).
At \(t=3\),
\(s=3+6(9)-27=30.\)
Since distance and displacement are equal at \(t=3\), the distance already travelled is \(30\text{ m}\).
The total distance is \(57\text{ m}\), so from \(t=3\) to \(t=k\), the additional distance is \(27\text{ m}\).
Using displacement for \(3\leqslant t\leqslant k\),
\(12k-\frac13k^2-3-30=27.\)
Thus
\(-\frac13k^2+12k-60=0.\)
Multiplying by \(-3\),
\(k^2-36k+180=0.\)
So
\((k-6)(k-30)=0.\)
The value \(k=30\) is not allowed because the velocity would be negative. Therefore
