<p>A particle moves along a straight line with an initial displacement at \(O\). Its velocity, \(\mathrm{v}\) m s\(^{-1}\), at a time \(t\) seconds is given by</p>
<p style="text-align: center">\(\mathrm{v}(t) = (3t^2 - 10t + 4)\mathrm{e}^{-0.5t} - 4\cos{t}\), where&nbsp;\(0 \leq t \leq 6\).</p>
<ol class="abc-list"><li data-list-item-id="item-1788761021436-ea58t411k" data-list-item-path="item-1788761021436-ea58t411k"><div class="list-item-paragraph"><p>Find the maximum displacement of the particle.</p><span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1788761021972-o29f05h33" data-list-item-path="item-1788761021972-o29f05h33"><div class="list-item-paragraph"><p>Determine the intervals of time when the acceleration of the particle is negative.</p><span class="inline-marks">[3]</span></div></li></ol>

A particle moves along a straight line with an initial displacement at \(O\). Its velocity, \(\mathrm{v}\) m s\(^{-1}\), at a time \(t\) seconds is given by

\(\mathrm{v}(t) = (3t^2 - 10t + 4)\mathrm{e}^{-0.5t} - 4\cos{t}\), where \(0 \leq t \leq 6\).

<p>Find the maximum displacement of the particle.</p>

<p>Determine the intervals of time when the acceleration of the particle is negative.</p>

Kinematics

International Baccalaureate Mathematics: Applications and Interpretation Standard Level

2026 May TZC P2 Q6

2026 May TZC P2 Q6