| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity-time graph |
| Difficulty | Easy -1.2 This is a straightforward mechanics question testing basic interpretation of velocity-time graphs. All parts require direct application of standard formulas: (a) area under graph for distance, (b) total distance/time for average speed, (c) gradient for acceleration, (d) F=ma. No problem-solving insight needed, just routine calculations from a clearly presented graph. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of triangle 0 to 10: \(\frac{1}{2} \times 10 \times 4 = 20\) m | M1 | Area of one section |
| Area of trapezium/triangle 10 to 30: \(\frac{1}{2}(10+20) \times 4 + \frac{1}{2} \times 10 \times 3 = ...\) or trapezium \(10\) to \(30\) then triangle \(30\) to \(40\) | M1 | Area of further section(s) |
| Segment 10–20: \(\frac{1}{2}(4+7)\times 10 = 55\) m | A1 | |
| Segment 20–30: \(\frac{1}{2}(7+7)\times 10 = 70\) m; Segment 30–40: \(\frac{1}{2}\times 10 \times 7 = 35\) m | A1 | Total = \(20 + 55 + 70 + 35 = 180\) m |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Average speed \(= \frac{\text{total distance}}{\text{total time}} = \frac{180}{40}\) | M1 | Their distance divided by 40 |
| \(= 4.5 \text{ m s}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a = \frac{v - u}{t} = \frac{4 - 0}{10}\) | M1 | |
| \(= 0.4 \text{ m s}^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F = ma = 200000 \times 0.4\) | M1 | Their \(a\) multiplied by \(200000\) kg |
| \(= 80000 \text{ N}\) (or \(80 \text{ kN}\)) | A1 |
# Question 2:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of triangle 0 to 10: $\frac{1}{2} \times 10 \times 4 = 20$ m | M1 | Area of one section |
| Area of trapezium/triangle 10 to 30: $\frac{1}{2}(10+20) \times 4 + \frac{1}{2} \times 10 \times 3 = ...$ or trapezium $10$ to $30$ then triangle $30$ to $40$ | M1 | Area of further section(s) |
| Segment 10–20: $\frac{1}{2}(4+7)\times 10 = 55$ m | A1 | |
| Segment 20–30: $\frac{1}{2}(7+7)\times 10 = 70$ m; Segment 30–40: $\frac{1}{2}\times 10 \times 7 = 35$ m | A1 | Total = $20 + 55 + 70 + 35 = 180$ m |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Average speed $= \frac{\text{total distance}}{\text{total time}} = \frac{180}{40}$ | M1 | Their distance divided by 40 |
| $= 4.5 \text{ m s}^{-1}$ | A1 | |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a = \frac{v - u}{t} = \frac{4 - 0}{10}$ | M1 | |
| $= 0.4 \text{ m s}^{-2}$ | A1 | |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F = ma = 200000 \times 0.4$ | M1 | Their $a$ multiplied by $200000$ kg |
| $= 80000 \text{ N}$ (or $80 \text{ kN}$) | A1 | |
---2 The graph shows how the velocity of a train varies as it moves along a straight railway line.\\
\includegraphics[max width=\textwidth, alt={}, center]{f30b02da-a41e-44cb-b45f-9e6a3a9d0528-04_574_1595_402_203}
\begin{enumerate}[label=(\alph*)]
\item Find the total distance travelled by the train.
\item Find the average speed of the train.
\item Find the acceleration of the train during the first 10 seconds of its motion.
\item The mass of the train is 200 tonnes. Find the magnitude of the resultant force acting on the train during the first 10 seconds of its motion.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2011 Q2 [10]}}