M2 June 2017 Q4
4. At time \(t = 0\) a particle \(P\) leaves the origin \(O\) and moves along the \(x\)-axis. At time \(t\) seconds, the velocity of \(P\) is \(v\) m s\(^{-1}\) in the positive \(x\) direction, where\[v = 3t^2 - 16t + 21\]The particle is instantaneously at rest when \(t = t_1\) and when \(t = t_2\) \((t_1 \lt t_2)\).
(a) Find the value of \(t_1\) and the value of \(t_2\). (2)
(b) Find the magnitude of the acceleration of \(P\) at the instant when \(t = t_1\). (3)
(c) Find the distance travelled by \(P\) in the interval \(t_1 \leqslant t \leqslant t_2\). (4)
(d) Show that \(P\) does not return to \(O\). (3)
| Scheme | Marks |
|---|---|
| \(v = 0 \Rightarrow 3t^2 - 16t + 21 = 0\) | M1 |
| \(\left((3t - 7)(t - 3) = 0\right)\ \ \ t_1 = \dfrac{7}{3},\ \ \ t_2 = 3\) | A1 |
| (2) |
Notes
M1 Set \(v = 0\) and attempt to solve
| Scheme | Marks |
|---|---|
| \(a = \dfrac{\mathrm{d}}{\mathrm{d}t}\left(3t^2 - 16t + 21\right)\) | M1 |
| \(= 6t - 16\) | A1 |
| \(t = t_1,\ \ \ a = 6 \times \dfrac{7}{3} - 16 = -2\) (m s\(^{-2}\)) Magnitude 2 (m s\(^{-2}\)) | A1 |
| (3) |
Notes
M1 Differentiate \(v\) to obtain \(a\)
A1 No errors seen. Must be positive - the Q asks for magnitude.
| Scheme | Marks |
|---|---|
| \(s = \displaystyle\int\left(3t^2 - 16t + 21\right)\mathrm{d}t\) | M1 |
| \(= t^3 - 8t^2 + 21t\ (+C)\) | A1 |
| \(\pm\left(\left(3^3 - 8 \times 9 + 21 \times 3\right) - \left(\left(\dfrac{7}{3}\right)^3 - 8 \times \dfrac{49}{9} + 21 \times \dfrac{7}{3}\right)\right)\) | M1 |
| \(s = 0.148\) (m) \(\ \ \left(\dfrac{4}{27}\right)\) | A1 |
| (4) |
Notes
M1 Integrate \(v\) to find \(s\)
M1 Correct use of their limits
A1 Final answer must be positive. 0.15 or better
| Scheme | Marks |
|---|---|
| Return to \(O \Rightarrow s = 0 = t\left(t^2 - 8t + 21\right)\) | B1 |
| Discriminant of quadratic \(= 64 - 4 \times 21\ (= -20) \lt 0\) | M1 |
| No real roots \(\Rightarrow\) does not return to \(O\) | A1 |
| (3) | |
| (12 marks) |
Notes
B1 seen or implied
M1 Or equivalent. *given answer so must show some evidence of method*
A1 Sufficient correct working to justify *given answer*
4dalt
| Travels away until \(t_1 = \dfrac{7}{3}\), turns back at \(t_2 = 3\) then turns away again | M1 |
| \(s_3 = 18\) | B1 |
| Complete argument | A1 |
M1 Complete story
B1 Seen or implied
4dalt
| Distance time graph | B1 |
| Locate min turning point | M1 |
| Complete argument | A1 |