| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find position by integrating velocity |
| Difficulty | Moderate -0.8 This is a straightforward integration question requiring integration of power functions (t and t^(1/2)) and application of initial conditions. The techniques are routine for M2 level with no problem-solving insight needed, making it easier than average but not trivial since it involves vector integration and distance calculation. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(r = t^2 i - \frac{2}{3}t^3 j - 10i + j = (t^2 - 10)i + (1 - \frac{2}{3}t^3)j\) | M1 M1 A1 A1 | |
| (b) When \(t = 4\), \(r = -6i - \frac{13}{3}j\) | M1 A1 | |
| \( | r | = 7.40 \text{ m}\) |
(a) $r = t^2 i - \frac{2}{3}t^3 j - 10i + j = (t^2 - 10)i + (1 - \frac{2}{3}t^3)j$ | M1 M1 A1 A1 |
(b) When $t = 4$, $r = -6i - \frac{13}{3}j$ | M1 A1 |
$|r| = 7.40 \text{ m}$ | A1 | 7 marks3. A particle $P$ moves in a horizontal plane such that, at time $t$ seconds, its velocity is $\mathbf { v } \mathrm { ms } ^ { - 1 }$, where $\mathbf { v } = 2 t \mathbf { i } - t ^ { \frac { 1 } { 2 } } \mathbf { j }$. When $t = 0 , P$ is at the point with position vector $- 10 \mathbf { i } + \mathbf { j }$ relative to a fixed origin $O$.
\begin{enumerate}[label=(\alph*)]
\item Find the position vector $\mathbf { r }$ of $P$ at time $t$ seconds.
\item Find the distance $O P$ when $t = 4$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q3 [7]}}