| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2002 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Vector motion with components |
| Difficulty | Moderate -0.3 Part (a) requires straightforward differentiation of velocity to show constant acceleration (2 marks of routine calculus). Part (b) involves integrating velocity to find position, applying initial conditions, then calculating distance using Pythagoras - standard M2 techniques with no novel insight required. The 6 marks reflect multiple steps rather than conceptual difficulty, making this slightly easier than average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02f Non-uniform acceleration: using differentiation and integration |
The velocity $v$ m s$^{-1}$ of a particle $P$ at time $t$ seconds is given by
$$\mathbf{v} = (3t - 2)\mathbf{i} - 5t\mathbf{j}.$$
\begin{enumerate}[label=(\alph*)]
\item Show that the acceleration of $P$ is constant. [2]
\end{enumerate}
At $t = 0$, the position vector of $P$ relative to a fixed origin O is $3\mathbf{i}$ m.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the distance of $P$ from O when $t = 2$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2002 Q1 [8]}}