| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Read and interpret velocity-time graph |
| Difficulty | Moderate -0.8 This is a straightforward M1 question combining basic kinematics with Newton's second law. Part (a) requires finding area under a velocity-time graph (trapezium), part (b) is simple gradient calculation, and part (c) applies F=ma with weight, all standard textbook exercises requiring only routine application of well-practiced techniques with no problem-solving insight needed. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| \(s = \frac{1}{2}(3+10) \times 3 = 19.5\) m | M1, A1, A1 | Finding distance by summing 3 areas or using formula for the area of a trapezium. Correct equation/3 correct expressions for the areas. Correct total distance. |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = \frac{3}{4} = 0.75\) ms\(^{-2}\) | B1 | Correct acceleration as a decimal or as a fraction |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - 400g = 400 \times 0.75\) | M1, A1F | Three term equation of motion containing \(T\), \(400g\) and \(400 \times 0.75\) or equivalent. Correct equation. |
| \(T = 3920 + 300 = 4220\) N | A1F | Correct tension. Only ft from \(a = \frac{4}{3}\) (ft 4453 N or 4450 N from \(a = \frac{4}{3}\) scores M1A1A1). M1 and A1 not dependent on B1 |
**1(a)**
| $s = \frac{1}{2}(3+10) \times 3 = 19.5$ m | M1, A1, A1 | Finding distance by summing 3 areas or using formula for the area of a trapezium. Correct equation/3 correct expressions for the areas. Correct total distance. |
**1(b)**
| $a = \frac{3}{4} = 0.75$ ms$^{-2}$ | B1 | Correct acceleration as a decimal or as a fraction |
**1(c)**
| $T - 400g = 400 \times 0.75$ | M1, A1F | Three term equation of motion containing $T$, $400g$ and $400 \times 0.75$ or equivalent. Correct equation. |
| $T = 3920 + 300 = 4220$ N | A1F | Correct tension. Only ft from $a = \frac{4}{3}$ (ft 4453 N or 4450 N from $a = \frac{4}{3}$ scores M1A1A1). M1 and A1 not dependent on B1 |1 The diagram shows a velocity-time graph for a lift.\\
\includegraphics[max width=\textwidth, alt={}, center]{a381686b-0b1e-41ba-b88f-be1601e42098-2_337_917_552_557}
\begin{enumerate}[label=(\alph*)]
\item Find the distance travelled by the lift.
\item Find the acceleration of the lift during the first 4 seconds of the motion.
\item The lift is raised by a single vertical cable. The mass of the lift is 400 kg . Find the tension in the cable during the first 4 seconds of the motion.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2008 Q1 [7]}}