| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Average speed calculation |
| Difficulty | Moderate -0.8 This is a straightforward application of SUVAT equations with clearly identified variables and standard methods. Parts (a) and (b) use s = (u+v)t/2 and v = u + at directly, while part (c) is another basic SUVAT calculation. No problem-solving insight required, just routine substitution into kinematic formulae—easier than average A-level questions. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(250 = \frac{1}{2}(17 + 33)t\) → \(t = 500 \div 50 = 10\) s | M1 A1 A1 | |
| (b) \(v = u + at\): \(33 = 17 + 10a\) → \(a = 1.6\) ms\(^{-2}\) | M1 A1 | |
| (c) \(s = \frac{1}{2}(33 + 0) \times 6 = 99\) m | M1 A1 | Total: 7 marks |
**(a)** $250 = \frac{1}{2}(17 + 33)t$ → $t = 500 \div 50 = 10$ s | M1 A1 A1 |
**(b)** $v = u + at$: $33 = 17 + 10a$ → $a = 1.6$ ms$^{-2}$ | M1 A1 |
**(c)** $s = \frac{1}{2}(33 + 0) \times 6 = 99$ m | M1 A1 | **Total: 7 marks**A body moves in a straight line with constant acceleration. Its speed increases from 17 ms$^{-1}$ to 33 ms$^{-1}$ while it travels a distance of 250 m. Find
\begin{enumerate}[label=(\alph*)]
\item the time taken to travel the 250 m, \hfill [3 marks]
\item the acceleration of the body. \hfill [2 marks]
\end{enumerate}
The body now decelerates at a constant rate from 33 ms$^{-1}$ to rest in 6 seconds.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the distance travelled in these 6 seconds. \hfill [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q3 [7]}}