Unit 29 Review Exercises

Kinematics

These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed.

1 Paper 1 A particle moves in a straight line with displacement \(s = t^3 - 9t^2 + 24t\) metres, \(t \ge 0\).

(a) Find \(v\) and \(a\).

(b) Find the times when the particle is at rest.

(c) Find the acceleration at the first of these times.

(a) \(v = 3t^2 - 18t + 24\), \(\;a = 6t - 18\).

(b) \(v = 3(t - 2)(t - 4) = 0 \Rightarrow t = 2\) or \(t = 4\).

(c) \(a(2) = -6\ \mathrm{m\,s^{-2}}\). See 29.1.

2 Paper 1 A train starts from rest. Its velocity is \(v = 0.5t\) for \(0 \le t \le 40\), \(\;v = 20\) for \(40 \le t \le 100\), and \(v = 70 - 0.5t\) for \(100 \le t \le 140\) (\(v\) in \(\mathrm{m\,s^{-1}}\), \(t\) in seconds).

(a) Find the acceleration in each stage.

(b) Find the total distance travelled.

(a) \(0.5\ \mathrm{m\,s^{-2}}\), \(\;0\), \(\;-0.5\ \mathrm{m\,s^{-2}}\).

(b) Area under the graph: \(\dfrac{1}{2}(40)(20) + 60 \times 20 + \dfrac{1}{2}(40)(20) = 400 + 1200 + 400 = 2000\) m. See 29.1 and 29.2.

3 Paper 1 A particle has displacement \(s = 5 + 4t - t^2\) metres, \(0 \le t \le 5\).

(a) Find its greatest displacement from \(O\).

(b) Find the total distance travelled from \(t = 0\) to \(t = 5\).

(a) \(v = 4 - 2t = 0\) at \(t = 2\); \(s(2) = 9\). Also \(s(0) = 5\) and \(s(5) = 0\), so the greatest displacement is \(9\) m.

(b) From \(s = 5\) to \(s = 9\), then back to \(s = 0\): distance \(= 4 + 9 = 13\) m. See 29.1.

4 Paper 1 A particle passes \(O\) at \(t = 0\) with velocity \(-12\ \mathrm{m\,s^{-1}}\). Its acceleration is \(a = 12t - 6\ \mathrm{m\,s^{-2}}\).

(a) Find \(v\) in terms of \(t\).

(b) Find the time when it comes to rest, and its displacement from \(O\) then.

(a) \(v = 6t^2 - 6t + c\), \(\;v(0) = -12 \Rightarrow v = 6t^2 - 6t - 12\).

(b) \(v = 6(t - 2)(t + 1) = 0 \Rightarrow t = 2\) (as \(t \ge 0\)). \(s = 2t^3 - 3t^2 - 12t\) (as \(s(0) = 0\)), so \(s(2) = 16 - 12 - 24 = -20\) m. See 29.2.

5 Paper 1 A particle has velocity \(v = 3t^2 - 18t + 24\ \mathrm{m\,s^{-1}}\), \(0 \le t \le 5\). Find the displacement and the total distance travelled from \(t = 0\) to \(t = 5\).

\(F(t) = t^3 - 9t^2 + 24t\). Displacement \(= F(5) - F(0) = 125 - 225 + 120 = 20\) m.

\(v = 3(t - 2)(t - 4)\). \(F(0) = 0\), \(F(2) = 20\), \(F(4) = 16\), \(F(5) = 20\). Distance \(= 20 + 4 + 4 = 28\) m. See 29.2.

6 Paper 1 A particle leaves \(O\) with velocity \(v = 4\mathrm{e}^{-0.5t}\ \mathrm{m\,s^{-1}}\). Find

(a) the initial acceleration, \(\;\)

(b) the displacement from \(O\) when \(t = 2\), in exact form.

(a) \(a = -2\mathrm{e}^{-0.5t}\), so \(a(0) = -2\ \mathrm{m\,s^{-2}}\).

(b) \(\displaystyle\int_0^2 4\mathrm{e}^{-0.5t}\,\mathrm{d}t = \Big[-8\mathrm{e}^{-0.5t}\Big]_0^2 = 8 - 8\mathrm{e}^{-1} = 8\left(1 - \mathrm{e}^{-1}\right)\) m. See 29.2.

7 Paper 1 A particle has displacement \(s = t^3 - 6t^2\) metres. Is it speeding up or slowing down

(a) when \(t = 1\), \(\;\) (b) when \(t = 3\)? Explain.

\(v = 3t^2 - 12t\), \(\;a = 6t - 12\).

(a) \(v(1) = -9\), \(a(1) = -6\): same sign, so speeding up.

(b) \(v(3) = -9\), \(a(3) = 6\): opposite signs, so slowing down. See 29.1.

8 Paper 2 A particle has velocity \(v = t\sin t - 1\ \mathrm{m\,s^{-1}}\), \(0 \le t \le 4\). Find

(a) the times when it changes direction, \(\;\)

(b) the displacement, \(\;\)

(c) the total distance travelled, from \(t = 0\) to \(t = 4\).

(a) Zeros of \(v\) on the GDC: \(t \approx 1.11\) s and \(t \approx 2.77\) s.

(b) \(\displaystyle\int_0^4 (t\sin t - 1)\,\mathrm{d}t \approx -2.14\) m.

(c) \(\displaystyle\int_0^4 \lvert t\sin t - 1 \rvert\,\mathrm{d}t \approx 3.91\) m. See 29.2.

9 Paper 1 Two particles move along the same line. Their displacements from \(O\) are \(s_A = t^2 - 2t + 4\) and \(s_B = 3t - 2\) metres, \(t \ge 0\). Find when they first meet, and decide whether they are moving in the same direction at that moment.

\(t^2 - 2t + 4 = 3t - 2 \Rightarrow t^2 - 5t + 6 = 0 \Rightarrow t = 2\) or \(t = 3\). They first meet at \(t = 2\).

\(v_A = 2t - 2\), so \(v_A(2) = 2\); \(\;v_B = 3\). Both positive: same direction. See 29.1 and 29.2.

10 Paper 1 A canoe stops paddling at \(t = 0\). Its velocity is then \(v = \dfrac{40}{(t + 2)^2}\ \mathrm{m\,s^{-1}}\).

(a) Show that the distance travelled in the first \(T\) seconds is \(20 - \dfrac{40}{T + 2}\) metres.

(b) Find how long it takes to travel \(15\) m.

(c) Explain why the canoe never travels \(20\) m.

(a) \(v > 0\), so distance \(= \displaystyle\int_0^T \frac{40}{(t + 2)^2}\,\mathrm{d}t = \left[-\frac{40}{t + 2}\right]_0^T = 20 - \frac{40}{T + 2}\).

(b) \(\dfrac{40}{T + 2} = 5 \Rightarrow T = 6\) s.

(c) \(\dfrac{40}{T + 2} > 0\) for every \(T\), so the distance is always less than \(20\) m. See 29.2.