| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Average speed or total distance calculation |
| Difficulty | Moderate -0.8 This is a straightforward multi-part SUVAT question where all necessary parameters are given explicitly for each stage. Students simply need to select and apply standard kinematic equations (s=½(u+v)t, v=u+at, s=ut+½at²) without any problem-solving insight or tricky setup. The calculations are routine and the question structure guides students through each step methodically. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(s = \frac{1}{2}(u+v)t\) | M1 | Use of correct suvat equation |
| \(640 = \frac{1}{2}(20+12)t\) | M1 | Correct substitution |
| \(t = 40\) seconds | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v^2 = u^2 + 2as\) or \(v = u + at\) | M1 | Use of correct suvat |
| \(144 = 400 + 2a(640)\) or \(12 = 20 + 40a\) | M1 | Correct substitution |
| \(a = -0.25 \text{ ms}^{-2}\), deceleration \(= 0.25 \text{ ms}^{-2}\) | A1 | Accept \(\frac{1}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(s = ut + \frac{1}{2}at^2\) | M1 | Use of correct suvat with \(u=12\), \(t=70\) |
| \(1820 = 12(70) + \frac{1}{2}a(70)^2\) | M1 | Correct substitution |
| \(a = \frac{2(1820-840)}{4900} = 0.4 \text{ ms}^{-2}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v = u + at\) | M1 | ft their acceleration |
| \(v = 12 + 0.4 \times 70\) | M1 | Correct substitution |
| \(v = 40 \text{ ms}^{-1}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Total time \(= 40 + 70 = 110\) seconds | M1 | Correct total time and distance |
| Average speed \(= \frac{2460}{110} = \frac{246}{11} \approx 22.4 \text{ ms}^{-1}\) | A1 | cao |
# Question 1:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s = \frac{1}{2}(u+v)t$ | M1 | Use of correct suvat equation |
| $640 = \frac{1}{2}(20+12)t$ | M1 | Correct substitution |
| $t = 40$ seconds | A1 | cao |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v^2 = u^2 + 2as$ or $v = u + at$ | M1 | Use of correct suvat |
| $144 = 400 + 2a(640)$ or $12 = 20 + 40a$ | M1 | Correct substitution |
| $a = -0.25 \text{ ms}^{-2}$, deceleration $= 0.25 \text{ ms}^{-2}$ | A1 | Accept $\frac{1}{4}$ |
## Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s = ut + \frac{1}{2}at^2$ | M1 | Use of correct suvat with $u=12$, $t=70$ |
| $1820 = 12(70) + \frac{1}{2}a(70)^2$ | M1 | Correct substitution |
| $a = \frac{2(1820-840)}{4900} = 0.4 \text{ ms}^{-2}$ | A1 | cao |
## Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v = u + at$ | M1 | ft their acceleration |
| $v = 12 + 0.4 \times 70$ | M1 | Correct substitution |
| $v = 40 \text{ ms}^{-1}$ | A1 | cao |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Total time $= 40 + 70 = 110$ seconds | M1 | Correct total time and distance |
| Average speed $= \frac{2460}{110} = \frac{246}{11} \approx 22.4 \text{ ms}^{-1}$ | A1 | cao |
---1 A car travels on a straight horizontal race track. The car decelerates uniformly from a speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to a speed of $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ as it travels a distance of 640 metres. The car then accelerates uniformly, travelling a further 1820 metres in 70 seconds.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the time that it takes the car to travel the first 640 metres.
\item Find the deceleration of the car during the first 640 metres.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find the acceleration of the car as it travels the further 1820 metres.
\item Find the speed of the car when it has completed the further 1820 metres.
\end{enumerate}\item Find the average speed of the car as it travels the 2460 metres.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2013 Q1 [14]}}