| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Particle motion: 2D constant acceleration |
| Difficulty | Moderate -0.8 This is a straightforward application of SUVAT equations in 2D vector form. Part (a) uses a = (v-u)/t directly, part (b) requires s = ut + ½at² or s = (u+v)t/2, and part (c) is simply total displacement divided by time. All three parts involve routine calculations with no problem-solving insight required, making this easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors3.02d Constant acceleration: SUVAT formulae3.02e Two-dimensional constant acceleration: with vectors |
**Question 4:**
- Velocity at A: $(4\mathbf{i} + 2\mathbf{j})$ m s$^{-1}$
- Velocity at B: $(7\mathbf{i} + 6\mathbf{j})$ m s$^{-1}$
- Time A→B: 10 seconds
**(a)** Acceleration [3 marks]
**(b)** Distance A to B [5 marks]
**(c)** Average velocity A to B [2 marks]
---4 A particle moves with constant acceleration between the points $A$ and $B$. At $A$, it has velocity $( 4 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$. At $B$, it has velocity $( 7 \mathbf { i } + 6 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$. It takes 10 seconds to move from $A$ to $B$.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the particle.
\item Find the distance between $A$ and $B$.
\item Find the average velocity as the particle moves from $A$ to $B$.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2015 Q4 [10]}}