M3 January 2012 Q3
3. A particle \(P\) is moving in a straight line. At time \(t\) seconds, \(P\) is at a distance \(x\) metres from a fixed point \(O\) on the line and is moving away from \(O\) with speed \(\dfrac{10}{x + 6}\) m s\(^{-1}\).
(a) Find the acceleration of \(P\) when \(x = 14\) (4)
Given that \(x = 2\) when \(t = 1\),
(b) find the value of \(t\) when \(x = 14\) (6)
| Scheme | Marks |
|---|---|
| \(a = v\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{10}{x + 6} \times \dfrac{-10}{(x + 6)^2},\ \ = \dfrac{-100}{(x + 6)^3}\) | M1 M1, A1 |
| \(= \dfrac{-100}{(14 + 6)^3} = -\dfrac{1}{80}\) ms\(^{-2}\) | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{10}{x + 6} \Rightarrow \displaystyle\int x + 6\,\mathrm{d}x = \int 10\,\mathrm{d}t\) | M1 M1 |
| \(\left[\dfrac{x^2}{2} + 6x\right]_2^{14} = \left[10t\right]_1^{T}\) | M1 A1 |
| \(\dfrac{196}{2} + 6 \times 14 - 2 - 12 = 10T - 10\) | M1 |
| \(178 = 10T \qquad T = 17.8\ (\text{s})\) | A1 |
| (6) | |
| (10 marks) |
Notes
(Corrected from the printed mark scheme: the last line is printed as \(178 = T\).)