AS June 2022 Q4
4. A particle \(P\) moves on the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v\ \text{m s}^{-1}\) in the direction of \(x\) increasing, where
\[v = \dfrac{1}{2}\left(3\mathrm{e}^{2t} - 1\right) \qquad t \geqslant 0\]The acceleration of \(P\) at time \(t\) seconds is \(a\ \text{m s}^{-2}\)
(a) Show that \(a = 2v + 1\) (2)
(b) Find the acceleration of \(P\) when \(t = 0\) (1)
(c) Find the exact distance travelled by \(P\) in accelerating from a speed of \(1\ \text{m s}^{-1}\) to a speed of \(4\ \text{m s}^{-1}\) (7)
| Scheme | Marks | AO |
|---|---|---|
| \(a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{1}{2} \times 6\mathrm{e}^{2t}\) | M1 | 1.1b |
| \(= 2v + 1\) * | A1* | 1.1b |
| (2) |
Notes
M1: Need to see evidence of attempt to differentiate \(v\) wrt \(t\), not just a statement of intent.
A1*: Given answer correctly obtained
| Scheme | Marks | AO |
|---|---|---|
| \(3\ (\text{m s}^{-2})\) | B1 | 1.1b |
| (1) |
Notes
B1: cao
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{1}{2}\left(3\mathrm{e}^{2t} - 1\right)\) and integrate | M1 | 3.3 |
| \(x = \dfrac{1}{2}\left(\dfrac{3}{2}\mathrm{e}^{2t} - t\right)(+C)\) | A1 | 1.1b |
| Put either \(\dfrac{1}{2}\left(3\mathrm{e}^{2t} - 1\right) = 1\) or 4 and solve for \(t\) | M1 | 2.1 |
| \(t = 0\) | A1 | 1.1b |
| \(t = \dfrac{1}{2}\ln 3\) \((0.549306\ldots)\) | A1 | 1.1b |
| Substitute their \(t\) values into their \(x\) expression and subtract | M1 | 3.1a |
| \(\dfrac{3}{2} - \dfrac{1}{4}\ln 3\) (m) | A1 | 1.1b |
| (7) | ||
| (10 marks) |
Notes
M1: Set up differential equation and attempt to solve
A1: Condone missing \(C\)
M1: Use at least one of the given speeds to find a \(t\) value
A1: cao
A1: 0.55 or better
M1: Substitute their \(t\) values to find \(x\) values and showing subtracting. Need to see evidence.
M0 if using 1 and 4.
A1: cao