M3 January 2007 Q1
1. A particle \(P\) moves along the \(x\)-axis. At time \(t = 0\), \(P\) passes through the origin \(O\), moving in the positive \(x\)-direction. At time \(t\) seconds, the velocity of \(P\) is \(v\) m s\(^{-1}\) and \(OP = x\) metres. The acceleration of \(P\) is \(\tfrac{1}{12}(30 - x)\) m s\(^{-2}\), measured in the positive \(x\)-direction.
Given that the maximum speed of \(P\) is 10 m s\(^{-1}\),
| Scheme | Marks |
|---|---|
| Maximum speed when accel. = 0 (o.e.) | B1 |
| (1) |
Notes
Allow “acceln > 0 for \(x < 30\), acceln < 0 for \(x > 30\)”
Also “accelerating for \(x < 30\), decelerating for \(x > 30\)”
But “acceln < 0 for \(x > 30\)” only is B0
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{12}(30 - x) = v\dfrac{\mathrm{d}v}{\mathrm{d}x}\) (acceln = … + attempt to integrate) | M1 |
| Use of \(v\dfrac{\mathrm{d}v}{\mathrm{d}x}\): \(\dfrac{v^2}{2} = \dfrac{1}{12}\left(30x - \dfrac{x^2}{2}\right)\ \ (+\,c)\) | ↓ M1 A1 |
| Substituting \(x = 30\), \(v = 10\) and finding \(c\) (= 12.5), or limits | ↓ M1 |
| \(v^2 = 25 + 5x - \tfrac{1}{12}x^2\) (o.e.) | A1 |
| (5) | |
| (6 marks) |
Notes
↓ marks a mark that depends on the M mark above it (an arrow in the scheme).
1st M1 will be generous for wrong form of acceln (e.g. \(\mathrm{d}v/\mathrm{d}x\))!
3rd M1 If use limits, they must use them in correct way with correct values
Final A1. Have to accept any expression, but it must be for \(v^2\) explicitly (not \(1/2v^2\)), and if in separate terms, one can expect like terms to be collected. Hence answer in form as above, or e.g. \(\tfrac{1}{12}\left(300 + 60x - x^2\right)\); also \(100 - \tfrac{1}{12}(30 - x)^2\)