| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding constants from motion conditions |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring differentiation of position vectors to find velocity and acceleration, then using the magnitude formula with a given speed to find a constant. All steps are routine applications of standard techniques with no conceptual challenges or novel problem-solving required. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}t^2 - 3t + 4 = 0\) | M1 | Set \(v = 0\) |
| \(t^2 - 6t + 8 = 0\) | ||
| \((t-2)(t-4) = 0\) | DM1 | Solve for \(v\) |
| \(t = 2\) s or \(4\) s | A1 A1 | |
| Total | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int \frac{1}{2}t^2 - 3t + 4 \, dt\) | M1 | Integration – majority of powers increasing |
| \(= \frac{1}{6}t^3 - \frac{3}{2}t^2 + 4t \ (+C)\) | A1 | Correct (\(+C\) not required) |
| \(s = \int_0^2 \frac{1}{2}t^2 - 3t + 4 \, dt - \int_2^4 \frac{1}{2}t^2 - 3t + 4 \, dt\) | DM1 | Correct strategy for finding the required distance. Follow their "2". Subtraction/swap limits/modulus signs |
| \(= \left[\frac{1}{6}t^3 - \frac{3}{2}t^2 + 4t\right]_0^2 - \left[\frac{1}{6}t^3 - \frac{3}{2}t^2 + 4t\right]_2^4\) | ||
| \(= \frac{8}{6} - 6 + 8 - \left(\frac{64}{6} - 24 + 16 - \left(\frac{8}{6} - 6 + 8\right)\right)\) | A1 | Correct unsimplified |
| \(= \frac{10}{3} - \frac{8}{3} + \frac{10}{3}\) | ||
| \(= 4\) | A1 | |
| Total | [9] |
## Question 3:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}t^2 - 3t + 4 = 0$ | M1 | Set $v = 0$ |
| $t^2 - 6t + 8 = 0$ | | |
| $(t-2)(t-4) = 0$ | DM1 | Solve for $v$ |
| $t = 2$ s or $4$ s | A1 A1 | |
| **Total** | **(4)** | |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \frac{1}{2}t^2 - 3t + 4 \, dt$ | M1 | Integration – majority of powers increasing |
| $= \frac{1}{6}t^3 - \frac{3}{2}t^2 + 4t \ (+C)$ | A1 | Correct ($+C$ not required) |
| $s = \int_0^2 \frac{1}{2}t^2 - 3t + 4 \, dt - \int_2^4 \frac{1}{2}t^2 - 3t + 4 \, dt$ | DM1 | Correct strategy for finding the required distance. Follow their "2". Subtraction/swap limits/modulus signs |
| $= \left[\frac{1}{6}t^3 - \frac{3}{2}t^2 + 4t\right]_0^2 - \left[\frac{1}{6}t^3 - \frac{3}{2}t^2 + 4t\right]_2^4$ | | |
| $= \frac{8}{6} - 6 + 8 - \left(\frac{64}{6} - 24 + 16 - \left(\frac{8}{6} - 6 + 8\right)\right)$ | A1 | Correct unsimplified |
| $= \frac{10}{3} - \frac{8}{3} + \frac{10}{3}$ | | |
| $= 4$ | A1 | |
| **Total** | **[9]** | |
---3.A particle $P$ moves in a horizontal plane.At time $t$ seconds,the position vector of $P$ is $\mathbf { r }$ metres relative to a fixed origin $O$ ,and $\mathbf { r }$ is given by
$$\mathbf { r } = \left( 18 t - 4 t ^ { 3 } \right) \mathbf { i } + c t ^ { 2 } \mathbf { j } ,$$
where $c$ is a positive constant.When $t = 1.5$ ,the speed of $P$ is $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ .Find
\begin{enumerate}[label=(\alph*)]
\item the value of $c$ ,
\item the acceleration of $P$ when $t = 1.5$ . $\mathbf { r }$ metres relative to a fixed origin $O$ ,and $\mathbf { r }$ is given by $$\begin{aligned} \mathbf { r } = \left( 18 t - 4 t ^ { 3 } \right) \mathbf { i } + c t ^ { 2 } \mathbf { j } , \\ \text { where } c \text { is a positive constant.When } t = 1.5 \text { ,the speed of } P \text { is } 15 \mathrm {~m} \mathrm {~s} ^ { - 1 } \text { .Find } \end{aligned}$$ (a)the value of $c$ ,
3.A particle $P$ moves in a horizontal plane.At time $t$ seconds,the position vector of $P$ is D墐\\
(b)the acceleration of $P$ when $t = 1.5$ .
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q3}}