June 2024 Paper 2 Q18
18 A particle is moving in a straight line through the origin \(O\)
The displacement of the particle, \(r\) metres, from \(O\), at time \(t\) seconds is given by
\[r = p + 2t - q\mathrm{e}^{-0.2t}\]where \(p\) and \(q\) are constants.
When \(t = 3\), the acceleration of the particle is \(-1.8\) m s−2
(a) Show that \(q \approx 82\) [5 marks]
(b) The particle has an initial displacement of 5 metres.
Find the value of \(p\)
Give your answer to two significant figures. [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses \(v = \dfrac{\mathrm{d}r}{\mathrm{d}t}\) to obtain an expression for \(v\) with one term correct PI by \(-0.04q\mathrm{e}^{-0.2t}\) | M1 | 3.4 |
| Obtains \(2 + 0.2q\mathrm{e}^{-0.2t}\) ACF PI by \(-0.04q\mathrm{e}^{-0.2t}\) | A1 | 1.1b |
| Obtains \(-0.04q\mathrm{e}^{-0.2t}\) ACF | A1 | 1.1b |
| Substitutes \(t\) = 3 and \(a = -1.8\) into their expression for \(a\) | M1 | 1.1a |
| Completes reasoned argument and concludes that \(q \approx 82\) Accept \(q\) rounds to 82 Must show either a correct expression for \(q\) or \(q\) = 81.9… CSO AG | R1 | 2.1 |
| (5) |
Typical solution
Using \(a = \dfrac{\mathrm{d}^2r}{\mathrm{d}t^2}\)
\[v = 2 + 0.2q\mathrm{e}^{-0.2t}\]\[a = -0.04q\mathrm{e}^{-0.2t}\]Using \(a = -1.8\) when \(t = 3\)
\[-1.8 = -0.04q\mathrm{e}^{-0.6}\]\[q = 81.995 \approx 82\]| Scheme | Marks | AO |
|---|---|---|
| Substitutes \(t\) = 0, \(r\) = 5 and their \(q\) into \(r = p + 2t - q\mathrm{e}^{-0.2t}\) | M1 | 3.4 |
| Obtains AWRT 87 | A1 | 1.1b |
| (2) | ||
| (7 marks) |