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June 2019 p42 q7
3829
Particles P and Q leave a fixed point A at the same time and travel in the same straight line. The velocity of P after t seconds is \(6t(t-3)\) m s-1 and the velocity of Q after t seconds is \((10 - 2t)\) m s-1.
Sketch, on the same axes, velocity-time graphs for P and Q for \(0 \leq t \leq 5\).
Verify that P and Q meet after 5 seconds.
Find the greatest distance between P and Q for \(0 \leq t \leq 5\).
Solution
(i) The velocity-time graph for P is a quadratic curve \(v = 6t(t-3) = 6t^2 - 18t\), which is a parabola opening upwards, passing through the origin \((0,0)\) and cutting the t-axis at \(t = 3\). The velocity-time graph for Q is a straight line \(v = 10 - 2t\), with a negative gradient, starting at \(v = 10\) when \(t = 0\) and reaching \(v = 0\) at \(t = 5\).
(ii) To verify that P and Q meet after 5 seconds, we find the displacement of each particle. For Q, the displacement is \(s = \int (10 - 2t) \, dt = 10t - t^2 + c\). Evaluating from \(t = 0\) to \(t = 5\), \(s(Q) = 25\) m. For P, the displacement is \(s = \int (6t^2 - 18t) \, dt = 2t^3 - 9t^2 + c\). Evaluating from \(t = 0\) to \(t = 5\), \(s(P) = 25\) m. Since both displacements are equal, P and Q meet at \(t = 5\).
(iii) The distance between P and Q is given by \(|s_P - s_Q| = |2t^3 - 8t^2 - 10t|\). To find the maximum distance, differentiate and solve \(6t^2 - 16t - 10 = 0\). Solving gives \(t = 3.19\). Evaluating the distance at this time gives the maximum distance \(PQ = 48.4\) m.
