M2 June 2015 Q6
6. A particle \(P\) moves on the positive \(x\)-axis. The velocity of \(P\) at time \(t\) seconds is \((2t^2 - 9t + 4)\) m s\(^{-1}\). When \(t = 0\), \(P\) is 15 m from the origin \(O\).
Find
| Scheme | Marks |
|---|---|
| At rest when \(v = 0\): \(\ (2t^2 - 9t + 4) = 0\) | M1 |
| \(= (2t - 1)(t - 4),\) | DM1 |
| \(t = \dfrac{1}{2},\ 4\) | A1 |
| (3) |
Notes
DM1 Solve for t. Dependent on the previous M1
A1 Incorrect answers with no method shown score M0A0
| Scheme | Marks |
|---|---|
| \(a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = 4t - 9\) | M1 A1 |
| \(t = 5,\ \ a = 11\) (m s\(^{-2}\)) | A1 |
| (3) |
Notes
M1 Differentiate \(v\) to obtain \(a\) (at least one power of \(t\) going down)
A1 Correct derivative
| Scheme | Marks |
|---|---|
| \(s = \displaystyle\int v\,\mathrm{d}t = \dfrac{2}{3}t^3 - \dfrac{9}{2}t^2 + 4t\ (+C)\) | M1 A1 |
| Use of \(t = 0,\ t = \dfrac{1}{2},\ t = 4,\ t = 5\) (and \(t = 0,\ s = 15\)) as limits in integrals | DM1 |
| \(\left[\dfrac{2}{3}t^3 - \dfrac{9}{2}t^2 + 4t(+15)\right]_0^{\frac{1}{2}} - \left[\dfrac{2}{3}t^3 - \dfrac{9}{2}t^2 + 4t(+15)\right]_{\frac{1}{2}}^4 + \left[\dfrac{2}{3}t^3 - \dfrac{9}{2}t^2 + 4t(+15)\right]_4^5\) | A1 |
| \(\left(0,\ \dfrac{23}{24},\ -\dfrac{40}{3},\ \dfrac{-55}{6}\right)\ \ \ = \dfrac{23}{24} + \dfrac{343}{24} + \dfrac{100}{24} = 19.4\) (m) \(\left(15,\ 15\dfrac{23}{24}\left(\dfrac{383}{24}\right),\ \dfrac{5}{3},\ 5.8\dot{3}\left(\dfrac{35}{6}\right)\right)\) | A1 |
| (5) | |
| (11 marks) |
Notes
M1 Integrate \(v\) to obtain \(s\) (at least one power of \(t\) going up)
DM1 Correct strategy for their limits - requires subtraction of the negative distance. Dependent on the previous M1 and at least one positive solution for \(t\) in (0,5) from (a)
A1 NB: \(\displaystyle\int_0^5 v\,\mathrm{d}t\) scores M0A0A0
A1 \(19\dfrac{5}{12}\ \left(\dfrac{233}{12}\right)\) or better