M2 June 2013 Q3
3. A particle \(P\) moves on the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v\) m s\(^{-1}\) in the direction of \(x\) increasing, where
\[v = 2t^2 - 14t + 20, \qquad t \geqslant 0\]Find
| Scheme | Marks |
|---|---|
| \(v = 0 = 2t^2 - 14t + 20\) | M1 |
| \(= 2(t - 2)(t - 5)\) | M1 |
| \(t = 2\) or \(t = 5\) | A1 |
| (3) |
Notes
M1 Set \(v = 0\)
M1 Solve for \(t\)
(The brackets in \(2(t - 2)(t - 5)\) are missing from the printed mark scheme.)
There are many different approaches to part (b). The allocation of the two M marks is
M1: A method to find the time when the velocity is a minimum
M1: Evaluate the speed at that time
| Scheme | Marks |
|---|---|
| \(t = 0,\ \ v = 20\) (m s\(^{-1}\)) | B1 |
| \(a = 4t - 14 = 0\) | M1 |
| \(t = \dfrac{7}{2},\ \ \ v = 2 \times \dfrac{3}{2} \times \dfrac{-3}{2} = \dfrac{-9}{2}\) | M1A1 |
| Max speed \(= 20\) ms\(^{-1}\) | A1 |
| (5) |
Notes
M1A1 Must see \(\pm 4.5\)
A1 Clearly stated & correct conclusion. Depends on the two M marks. From correct solution only.
b alt1
| \(t = 0,\ v = 20\) (m s\(^{-1}\)) | B1 |
| Sketch with symmetry about their \(t = 3.5\) | M1 |
| \(v\)(their 3.5) | M1 |
| -4.5 | A1 |
| Max speed \(= 20\) ms\(^{-1}\) | A1 |
M1 Evaluate \(v\) at min.
A1 Correct work
A1 Clearly stated & correct conclusion. Depends on the two M marks. From correct solution only.
b alt 2
| \(t = 0,\ \ v = 20\) (m s\(^{-1}\)) | B1 |
| Justification of minimum or tabulate sufficient values to confirm location | M1 |
| Evaluate \(v\) at min. | M1 |
| Correct work | A1 |
| Correct conclusion. Depends on the two M marks | A1 |
A1 Clearly stated & from correct solution only.
b alt 3
| \(t = 0,\ \ v = 20\) (m s\(^{-1}\)) | B1 |
| Complete the square as far as \(\left(t - \dfrac{7}{2}\right)^2\) | M1 |
| \(2\left(t - \dfrac{7}{2}\right)^2 - \dfrac{9}{2}\) | M1A1 |
| Max speed \(= 20\) ms\(^{-1}\) | A1 |
A1 Clearly stated & correct conclusion. Depends on the two M marks. From correct solution only.
| Scheme | Marks |
|---|---|
| \(\displaystyle\int 2t^2 - 14t + 20\ \mathrm{d}t = \dfrac{2}{3}t^3 - 7t^2 + 20t\ (+C)\) | M1 A1 |
| Distance \(= \left[\dfrac{2}{3}t^3 - 7t^2 + 20t\right]_0^2 - \left[\dfrac{2}{3}t^3 - 7t^2 + 20t\right]_2^4\) | M1 A1 |
| \(= 2 \times \left[\dfrac{2}{3}t^3 - 7t^2 + 20t\right]^2 - \left[\dfrac{2}{3}t^3 - 7t^2 + 20t\right]_4\) | |
| \(= 2\left[\dfrac{16}{3} - 7 \times 4 + 40\right] - \left[\dfrac{2 \times 64}{3} - 7 \times 16 + 80\right] = 24\) (m) | A1 |
| (5) | |
| (13 marks) |
Notes
M1 Integration. Need to see majority of powers going up
A1 All correct. Condone \(C\) missing
M1 Correct method to find the distance, for their 2
A1 Correct unsimplified