| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Find acceleration from distances/times |
| Difficulty | Moderate -0.8 This is a straightforward three-part SUVAT question with clearly defined stages and standard application of kinematic equations. Part (a) uses s=ut+½at² with known values, part (b) uses v²=u²+2as, and part (c) uses s=ut+½at² again. All values are given explicitly, requiring only systematic substitution into standard formulae with no problem-solving insight or geometric reasoning needed. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(AB: 50 = 2 \times 22.5 + \frac{1}{2}a.4 \Rightarrow a = 2.5 \text{ m s}^{-2}\) | M1 A1 | (3) |
| (b) \(v^2 = 22.5^2 + 2 \times 2.5 \times 100 \Rightarrow v \approx 31.7(2) \text{ m s}^{-1}\) | M1 A1∇ | A1 (3) |
| (c) \(v_B = 22.5 + 2 \times 2.5 = 27.5\) (must be used) | M1 | |
| \(31.72 = 27.5 + 2.5t\) OR \(50 = 27.5t + \frac{1}{2} \times 2.5t^2\) OR \(50 = \frac{1}{2}(27.5 + 31.72)t \Rightarrow t \approx 1.69 \text{ s}\) | M1 A1∇ | A1 (4) |
| OR \(31.72 = 22.5 + 2.5T\) OR \(100 = 22.5t + \frac{1}{2} \times 2.5T^2 \Rightarrow T \approx 3.69 \Rightarrow t \approx 3.69 - 2 = 1.69 \text{ s}\) | M1 A1∇ | (4) |
| OR \(50 = 31.7t - \frac{1}{2} \times 2.5t^2\) Solve quadratic to get \(t = 1.69 \text{ s}\) | M2 A1∇ | A1 (4) |
| (a) $AB: 50 = 2 \times 22.5 + \frac{1}{2}a.4 \Rightarrow a = 2.5 \text{ m s}^{-2}$ | M1 A1 | (3) |
| (b) $v^2 = 22.5^2 + 2 \times 2.5 \times 100 \Rightarrow v \approx 31.7(2) \text{ m s}^{-1}$ | M1 A1∇ | A1 (3) |
| (c) $v_B = 22.5 + 2 \times 2.5 = 27.5$ (must be used) | M1 | |
| $31.72 = 27.5 + 2.5t$ OR $50 = 27.5t + \frac{1}{2} \times 2.5t^2$ OR $50 = \frac{1}{2}(27.5 + 31.72)t \Rightarrow t \approx 1.69 \text{ s}$ | M1 A1∇ | A1 (4) |
| **OR** $31.72 = 22.5 + 2.5T$ OR $100 = 22.5t + \frac{1}{2} \times 2.5T^2 \Rightarrow T \approx 3.69 \Rightarrow t \approx 3.69 - 2 = 1.69 \text{ s}$ | M1 A1∇ | (4) |
| **OR** $50 = 31.7t - \frac{1}{2} \times 2.5t^2$ Solve quadratic to get $t = 1.69 \text{ s}$ | M2 A1∇ | A1 (4) |
**Guidance:** NB note slight changes to scheme: dependency now in (c) and new rule on accuracy of answers. (b) M1 for valid use of data (e.g. finding speed at $B$ by spurious means and using this to get $v$ at $C$ is M0). Accept answer as AWRT 31.7. In (b) and (c), f.t. A marks are for f.t. on wrong $a$ and/or answer from (b). (c) M1 + M1 to get to an equation in the required $t$ (normally two stages, but they can do it in one via 3rd alternative above). Answer is cao. Hence premature approx (→ e.g. 1.68) is A0. But if they use a 3 s.f. answer from (b) and then give answer to (c) as 1.7, allow full marks. And accept 2 or 3 s.f. answer or better to (c).
---3. A train moves along a straight track with constant acceleration. Three telegraph poles are set at equal intervals beside the track at points $A , B$ and $C$, where $A B = 50 \mathrm {~m}$ and $B C = 50 \mathrm {~m}$. The front of the train passes $A$ with speed $22.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and 2 s later it passes $B$. Find
\begin{enumerate}[label=(\alph*)]
\item the acceleration of the train,
\item the speed of the front of the train when it passes $C$,
\item the time that elapses from the instant the front of the train passes $B$ to the instant it passes $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2006 Q3 [10]}}