| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Time to reach midpoint or specific position |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT problem requiring direct application of standard kinematic equations. Part (a) uses s = ut + ½at² or v² = u² + 2as to find acceleration, while part (b) applies the same equations to half the distance. Both parts involve routine algebraic manipulation with no conceptual challenges or problem-solving insight required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(s = vt - \frac{1}{2}at^2\) | M1 A2 | First M1 for complete method to produce value for \(a\); A2 if all correct, A1A0 for one error |
| \(40 = 10 \times 5 - \frac{1}{2}a(5)^2\) | Possible equations: \(40 = 5u + \frac{1}{2}a(5)^2\); \(10^2 = u^2 + 2a(40)\); \(10 = u + 5a\); \(40 = \frac{(u+10)}{2} \cdot 5\) | |
| \(a = 0.8\) | A1 (4) | Third A1 for \(0.8\) (m s\(^{-2}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Finding \(u\) \((= 6)\) | M1 | First M1 for attempt to find \(u\) (may have been done in (a) but MUST be used in (b)) |
| \(s = ut + \frac{1}{2}at^2\) (A to M) | M1 | Second M1 for complete method (may involve 2 or more suvat equations) for finding equation in \(t\) only |
| \(20 = 6t + \frac{1}{2}(0.8)t^2\) | A1 | First A1 for correct equation |
| \(t = \frac{-15 \pm \sqrt{225 + 200}}{2}\) | DM1 | Third M1 dependent on previous M, for solving equation for \(t\) |
| \(t = 2.8\) or \(2.81\) or better | A1 (5) | Second A1 for \(2.8\) s or better: \(\frac{5(2\sqrt{17}-6)}{4}\); \(\frac{40}{6+2\sqrt{17}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Finding \(v\) \((= \sqrt{68})\) | M1 | |
| \(s = vt - \frac{1}{2}at^2\) (A to M) | M1 | |
| \(20 = \sqrt{68}\,t - \frac{1}{2}(0.8)t^2\) | A1 | |
| \(t = \frac{\sqrt{68} \pm \sqrt{68 - 32}}{0.8}\) | DM1 | |
| \(t = 2.8\) or \(2.81\) or better | A1 (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(s = vt_1 - \frac{1}{2}at_1^2\) (M to B) | M2 | |
| \(20 = 10t_1 - \frac{1}{2}(0.8)t_1^2\) | A1 | |
| \(t_1 = \frac{10 \pm \sqrt{100 - 32}}{0.8}\) | DM1 | |
| \(t_1 = 2.192\) | A1 | |
| \(t = 5 - t_1 = 2.8\) or \(2.81\) or better | (5) |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s = vt - \frac{1}{2}at^2$ | M1 A2 | First M1 for complete method to produce value for $a$; A2 if all correct, A1A0 for one error |
| $40 = 10 \times 5 - \frac{1}{2}a(5)^2$ | | Possible equations: $40 = 5u + \frac{1}{2}a(5)^2$; $10^2 = u^2 + 2a(40)$; $10 = u + 5a$; $40 = \frac{(u+10)}{2} \cdot 5$ |
| $a = 0.8$ | A1 (4) | Third A1 for $0.8$ (m s$^{-2}$) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Finding $u$ $(= 6)$ | M1 | First M1 for attempt to find $u$ (may have been done in (a) but MUST be used in (b)) |
| $s = ut + \frac{1}{2}at^2$ (A to M) | M1 | Second M1 for complete method (may involve 2 or more suvat equations) for finding equation in $t$ only |
| $20 = 6t + \frac{1}{2}(0.8)t^2$ | A1 | First A1 for correct equation |
| $t = \frac{-15 \pm \sqrt{225 + 200}}{2}$ | DM1 | Third M1 dependent on previous M, for solving equation for $t$ |
| $t = 2.8$ or $2.81$ or better | A1 (5) | Second A1 for $2.8$ s or better: $\frac{5(2\sqrt{17}-6)}{4}$; $\frac{40}{6+2\sqrt{17}}$ |
**Alternative 1:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Finding $v$ $(= \sqrt{68})$ | M1 | |
| $s = vt - \frac{1}{2}at^2$ (A to M) | M1 | |
| $20 = \sqrt{68}\,t - \frac{1}{2}(0.8)t^2$ | A1 | |
| $t = \frac{\sqrt{68} \pm \sqrt{68 - 32}}{0.8}$ | DM1 | |
| $t = 2.8$ or $2.81$ or better | A1 (5) | |
**Alternative 2:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s = vt_1 - \frac{1}{2}at_1^2$ (M to B) | M2 | |
| $20 = 10t_1 - \frac{1}{2}(0.8)t_1^2$ | A1 | |
| $t_1 = \frac{10 \pm \sqrt{100 - 32}}{0.8}$ | DM1 | |
| $t_1 = 2.192$ | A1 | |
| $t = 5 - t_1 = 2.8$ or $2.81$ or better | (5) | |
---6. A cyclist is moving along a straight horizontal road and passes a point $A$. Five seconds later, at the instant when she is moving with speed $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, she passes the point $B$. She moves with constant acceleration from $A$ to $B$.
Given that $A B = 40 \mathrm {~m}$, find
\begin{enumerate}[label=(\alph*)]
\item the acceleration of the cyclist as she moves from $A$ to $B$,
\item the time it takes her to travel from $A$ to the midpoint of $A B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2017 Q6 [9]}}