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Nov 2009 p41 q7
3785
A particle P starts from rest at the point A at time t = 0, where t is in seconds, and moves in a straight line with constant acceleration a m s-2 for 10 s. For 10 ≤ t ≤ 20, P continues to move along the line with velocity v m s-1, where v = \(\frac{800}{t^2} - 2\). Find
the speed of P when t = 10, and the value of a,
the value of t for which the acceleration of P is -a m s-2,
the displacement of P from A when t = 20.
Solution
(i) For 0 ≤ t ≤ 10, the particle moves with constant acceleration a. At t = 10, the velocity is given by the velocity function: \(v = \frac{800}{10^2} - 2 = 8 - 2 = 6\) m s-1. Therefore, the speed of P when t = 10 is 6 m s-1. Since the particle starts from rest, \(v = at\), so \(a = \frac{6}{10} = 0.6\) m s-2.
(ii) For 10 ≤ t ≤ 20, the acceleration is given by differentiating the velocity function: \(a(t) = \frac{d}{dt} \left( \frac{800}{t^2} - 2 \right) = -\frac{1600}{t^3}\). Set \(a(t) = -0.6\) to find \(t\): \(-\frac{1600}{t^3} = -0.6\) implies \(t^3 = \frac{1600}{0.6}\), so \(t = 13.9\).
(iii) To find the displacement from A when t = 20, integrate the velocity function: \(s = \int (\frac{800}{t^2} - 2) \, dt = -\frac{800}{t} - 2t + C\). Using \(s(10) = 30\) to find \(C\), we have \(-\frac{800}{10} - 20 + C = 30\), so \(C = 130\). Then, \(s(20) = -\frac{800}{20} - 40 + 130 = 50\). Therefore, the displacement from A is 50 m.
