| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Constant acceleration vector (i and j) |
| Difficulty | Moderate -0.5 This is a straightforward vector mechanics question requiring routine differentiation of velocity to find acceleration, application of F=ma, and magnitude calculation. All steps are standard M2 techniques with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation3.01b Derived quantities and units3.02a Kinematics language: position, displacement, velocity, acceleration3.03c Newton's second law: F=ma one dimension |
1 A particle moves under the action of a force, $\mathbf { F }$ newtons. At time $t$ seconds, the velocity, $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$, of the particle is given by
$$\mathbf { v } = \left( t ^ { 3 } - 15 t - 5 \right) \mathbf { i } + \left( 6 t - t ^ { 2 } \right) \mathbf { j }$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the acceleration of the particle at time $t$.
\item The mass of the particle is 4 kg .
\begin{enumerate}[label=(\roman*)]
\item Show that, at time $t$,
$$\mathbf { F } = \left( 12 t ^ { 2 } - 60 \right) \mathbf { i } + ( 24 - 8 t ) \mathbf { j }$$
\item Find the magnitude of $\mathbf { F }$ when $t = 2$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M2 2009 Q1 [9]}}