A2 June 2019 Paper 1 Q17
17 A cyclist accelerates from rest for 5 seconds then brakes for 5 seconds, coming to rest at the end of the 10 seconds. The total mass of the cycle and rider is \(m\) kg, and at time \(t\) seconds, for \(0 \leqslant t \leqslant 10\), the cyclist’s velocity is \(v\,\mathrm{m\,s^{-1}}\).
A resistance to motion, modelled by a force of magnitude \(0.1mv\) N, acts on the cyclist during the whole 10 seconds.
During the braking phase of the motion, for \(5 \leqslant t \leqslant 10\), the brakes apply an additional constant resistance force of magnitude \(2m\) N and the cyclist does not provide any driving force.
During the acceleration phase \((0 \leqslant t \leqslant 5)\), the cyclist applies a driving force of magnitude directly proportional to \(t\).
| Scheme | Marks | AO |
|---|---|---|
| The resistance force is likely to increase with velocity. | B1 | 3.5b |
| [1] |
Notes
B1: allow ‘proportional to’, ‘varies with’
| Scheme | Marks | AO |
|---|---|---|
| \(m\dfrac{\mathrm{d}v}{\mathrm{d}t} = -2m - 0.1mv \Rightarrow \dfrac{\mathrm{d}v}{\mathrm{d}t} + 0.1v = -2\) | B1 | 3.3 |
| [1] |
Notes
B1: [by Newton’s 2nd Law]
| Scheme | Marks | AO |
|---|---|---|
| (i) IF \(\mathrm{e}^{0.1t}\) | M1 | 1.1a |
| \(\Rightarrow \dfrac{\mathrm{d}}{\mathrm{d}t}\left(v\mathrm{e}^{0.1t}\right) = -2\mathrm{e}^{0.1t}\) | M1 | 1.1b |
| \(\displaystyle\Rightarrow v\mathrm{e}^{0.1t} = \int -2\mathrm{e}^{0.1t}\,\mathrm{d}t = -20\mathrm{e}^{0.1t} + c\) | A1 | 1.1b |
| when \(t = 10\), \(v = 0 \Rightarrow c = 20\mathrm{e}\) | M1 | 3.1b |
| \(\Rightarrow v = 20(\mathrm{e}^{1 - 0.1t} - 1)\) | A1cao | 3.4 |
| [5] | ||
| (ii) When \(t = 5\), \(v = 20(\mathrm{e}^{0.5} - 1) = 12.97\ \mathrm{m\,s^{-1}}\). | B1 | 3.5a |
| [1] |
Notes
(c)(i)
M1: must be correct
M1: substituting \(t = 10\), \(v = 0\)
A1cao: or \(54.4\mathrm{e}^{-0.1t} - 20\)
Alternative solution
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{\mathrm{d}v}{2 + 0.1v} = -\int \mathrm{d}t\) | M1 |
| \(\Rightarrow 10\ln(2 + 0.1v) = -t + c\) | A1 |
| When \(t = 10\), \(v = 0 \Rightarrow c = 10\ln 2 + 10\) | M1 |
| \(\Rightarrow \ln(2 + 0.1v) = \ln 2 + 1 - 0.1t\) \(2 + 0.1v = \mathrm{e}^{\ln 2 + 1 - 0.1t} = 2\mathrm{e}^{1 - 0.1t}\) | M1 |
| \(\Rightarrow v = 20(\mathrm{e}^{1 - 0.1t} - 1)\) | A1cao |
| [5] |
M1: substituting \(t = 10\), \(v = 0\); M1: anti-logging
Alternative solution
| Scheme | Marks |
|---|---|
| AE \(\lambda + 0.1 = 0 \Rightarrow\) cf \(v = A\mathrm{e}^{-0.1t}\) | M1 |
| PI \(v = k \Rightarrow k = -20\) | B1 |
| GS \(v = A\mathrm{e}^{-0.1t} - 20\) | A1 |
| When \(t = 10\), \(v = 0\): \(0 = A\mathrm{e}^{-1} - 20 \Rightarrow A = 20\mathrm{e}\) | M1 |
| \(\Rightarrow v = 20(\mathrm{e}^{1 - 0.1t} - 1)\) | A1cao |
| [5] |
M1: substituting \(t = 10\), \(v = 0\)
(corrected from the printed mark scheme: the printed line reads “When \(t = 0\). \(v = 10\)”; the condition used is \(t = 10\), \(v = 0\).)
(c)(ii)
B1: 13 or better
| Scheme | Marks | AO |
|---|---|---|
| \(m\dfrac{\mathrm{d}v}{\mathrm{d}t} = ct - 0.1mv \Rightarrow \dfrac{\mathrm{d}v}{\mathrm{d}t} + 0.1v = \lambda t\) where \(\lambda = \dfrac{c}{m}\) | B1 | 3.3 |
| [1] |
Notes
B1: [by Newton’s 2nd Law]
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(v\mathrm{e}^{0.1t}\right) = \lambda t\mathrm{e}^{0.1t}\) | M1 | 2.1 |
| \(\displaystyle\Rightarrow v\mathrm{e}^{0.1t} = \int \lambda t\mathrm{e}^{0.1t}\,\mathrm{d}t = 10\lambda t\mathrm{e}^{0.1t} - \int 10\lambda\mathrm{e}^{0.1t}\,\mathrm{d}t\) | M1 | 2.1 |
| \(\Rightarrow v\mathrm{e}^{0.1t} = 10\lambda t\mathrm{e}^{0.1t} - 100\lambda\mathrm{e}^{0.1t} + c\) | A1 | 2.1 |
| When \(t = 0\), \(v = 0 \Rightarrow c = 100\lambda\) | M1 | 3.1b |
| \(\Rightarrow v = 10\lambda(t - 10 + 10\mathrm{e}^{-0.1t})\,*\) | A1 | 2.1 |
| [5] | ||
| (ii) When \(t = 5\), \(20(\mathrm{e}^{0.5} - 1) = 10\lambda(10\mathrm{e}^{-0.5} - 5)\) | M1 | 3.1b |
| \(\Rightarrow \lambda = 1.218\) | A1 | 1.1b |
| [2] |
Notes
(e)(i)
M1: integrating by parts
M1: substituting \(t = 0\), \(v = 0\)
A1: NB AG
Alternative solution
| Scheme | Marks |
|---|---|
| CF \(v = A\mathrm{e}^{-0.1t}\) | M1 |
| PI \(v = Ct + D\) \(C + 0.1(Ct + D) = \lambda t \Rightarrow C = 10\lambda,\ D = -100\lambda\) | M1 A1 |
| GS \(v = A\mathrm{e}^{-0.1t} + 10\lambda t - 100\lambda\) \(0 = A - 100\lambda \Rightarrow A = 100\lambda\) | M1 |
| \(v = 10\lambda(t - 10 + 10\mathrm{e}^{-0.1t})\,*\) | A1cao |
| [5] |
M1: substituting \(t = 0\), \(v = 0\); A1cao: NB AG
(e)(ii)
M1: subst \(t = 5\) and equating to their \(v\) when \(t = 5\)
A1: 1.2 or better
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle s_1 = \int_0^5 10\lambda(t - 10 + 10\mathrm{e}^{-0.1t})\,\mathrm{d}t\) | M1 | 3.1b |
| \(= 10\lambda\left[\dfrac{1}{2}t^2 - 10t - 100\mathrm{e}^{-0.1t}\right]_0^5\) | B1 | 1.1b |
| \(= 12.18(12.5 + 50 - 100\mathrm{e}^{-0.5}) = 22.49\) (m) | A1 | 1.1b |
| \(\displaystyle s_2 = \int_5^{10} 20(\mathrm{e}^{1 - 0.1t} - 1)\,\mathrm{d}t\) \(= 20\left[-10\mathrm{e}^{1 - 0.1t} - t\right]_5^{10}\) | M1 | 3.1b |
| \(= 20(-15 + 10\mathrm{e}^{0.5}) = 29.74\) (m) | A1 | 1.1b |
| Total distance = 52 m | A1cao | 3.2b |
| [6] |
Notes
M1: integrating \(v\) between 0, 5
B1: \(\left[\frac{1}{2}t^2 - 10t - 100\mathrm{e}^{-0.1t}\right]\)
A1: art 22.5 (soi)
M1: integrating their \(v\) between 5 and 10
A1: art 29.7 (soi)