M3 June 2013 (R) Q4
4. A particle \(P\) is moving along the positive \(x\)-axis. At time \(t\) seconds, \(t \geqslant 0\), \(P\) is \(x\) metres from the origin \(O\) and is moving away from \(O\) with velocity \(v\) m s\(^{-1}\), where \(v = \dfrac{4}{(x + 2)}\). When \(t = 0\), \(P\) is at \(O\). Find
(a) the distance of \(P\) from \(O\) when \(t = 2\) (5)
(b) the magnitude and direction of the acceleration of \(P\) when \(t = 2\) (5)
| Scheme | Marks |
|---|---|
| \(v = \dfrac{4}{(x + 2)} = \dfrac{\mathrm{d}x}{\mathrm{d}t}\) | B1 |
| \(\dfrac{\mathrm{d}t}{\mathrm{d}x} = \dfrac{x + 2}{4};\ \ \displaystyle\int_{t=0}^{t=2} 1\,\mathrm{d}t = \frac{1}{4}\int_{x=0}^{x=X} (x + 2)\,\mathrm{d}x,\ \ [t]_0^2 = \frac{1}{4}\left[\frac{x^2}{2} + 2x\right]_0^X\) | M1,A1 |
| \(2 = \dfrac{X^2}{8} + \dfrac{X}{2},\) | |
| \(0 = X^2 + 4X - 16,\ \ \ \ X = \dfrac{-4 + \sqrt{80}}{2} = 2.47\) (m) | M1depA1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(a\left(= \dfrac{\mathrm{d}v}{\mathrm{d}t}\right) = v\dfrac{\mathrm{d}v}{\mathrm{d}x}\) | B1 |
| \(= \dfrac{4}{(x + 2)} \times \dfrac{-4}{(x + 2)^2}\) | M1A1 |
| \(= \dfrac{-16}{(2.47 + 2)^3} = -0.1788\ldots\) their \(X\) | M1dep |
| 0.18 (m s\(^{-2}\)) towards \(O\). | A1 |
| (5) | |
| (10 marks) |