| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | Specimen |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Two vehicles: overtaking or meeting (algebraic) |
| Difficulty | Moderate -0.8 This is a standard M1 two-particle overtaking problem with straightforward SUVAT applications. Part (a) is direct use of v=u+at, parts (b) and (e) are routine graph sketching, and parts (c)-(d) require equating distances using standard kinematic equations. All steps are textbook-standard with no novel problem-solving required, making it easier than average but not trivial due to the multi-part nature and need for careful algebraic manipulation. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = u + at \Rightarrow 14 = 3.5a\) | M1 | Use of *suvat* to form an equation in \(a\) |
| \(a = 4\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Graph for \(A\) or \(B\) | B1 | |
| Second graph correct and both graphs extending beyond the point of intersection | B1 | |
| Values 3.5, 14, \(T\) shown on axes, with \(T\) not at the point of intersection | B1 | Accept labels with delineators |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}T \cdot 3T\), \(\quad \frac{(T+T-3.5)}{2} \cdot 14\) | M1 | Find distance for A or B in terms of \(T\) only. Correct area formulae: must see \(\frac{1}{2}\) in area formula and be adding in trapezium |
| One distance correct | A1 | |
| Both distances correct | A1 | |
| \(\frac{1}{2}T \cdot 3T = \frac{(T+T-3.5)}{2} \cdot 14\) | M1 | Equate distances and simplify to a 3 term quadratic in \(T\) in the form \(aT^2 + bT + c = 0\) |
| \(\frac{1}{2}T \cdot 3T = \frac{1}{2} \times 4 \times 3.5^2 + 14(T-3.5)\) | ||
| \(3T^2 - 28T + 49 = 0\) | A1 | Correct quadratic |
| \((3T-7)(T-7) = 0\) | M1 | Solve 3 term quadratic for \(T\) |
| \(T = \frac{7}{3}\) or \(7\) | A1 | Correct solution(s) - can be implied if only ever see \(T=7\) from correct work |
| but \(T > 3.5\), \(\quad T = 7\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 73.5 m | B1 | From correct work only. B0 if extra answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Graph showing \(A\) at velocity 4 (horizontal line) | B1 | Condone missing 4 |
| Graph showing \(B\) (must go beyond 3.5, condone no 3) | B1 | Condone graph going beyond \(T=7\) |
| Graph for \(A\) going beyond \(T=7\), dashed line shown | B1 | Condone graph going beyond \(T=7\). B0 if see a solid vertical line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Graphs for \(A\) and \(B\) | B1, B1, B0 | B1B1B0 on graph then max 5/8 for (c) if they do not solve for \(T\) in the question |
| \(\frac{1}{2} \times 3 \times (T+3.5)^2 = \frac{1}{2} \times 4 \times 3.5^2 + 14T\) | M1 | Use diagram to find area |
| One distance correct | A1 | |
| Both distances correct | A1 | |
| \(12T^2 - 28T - 49 = 0\) | M1 | Simplify to a 3 term quadratic in \(T\) |
| Correct quadratic | A1 | |
| \((2T-7)(6T+7) = 0\) | M1 | Complete method to solve for \(T\) |
| \(T = \frac{7}{2}\) or \(\frac{-7}{6}\) | A1 | Correct solution(s) - implied if only see Total \(= 7\) |
| Total time \(= 7\) | A1 |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = u + at \Rightarrow 14 = 3.5a$ | M1 | Use of *suvat* to form an equation in $a$ |
| $a = 4$ | A1 | |
**Total: (2)**
---
## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Graph for $A$ or $B$ | B1 | |
| Second graph correct and both graphs extending beyond the point of intersection | B1 | |
| Values 3.5, 14, $T$ shown on axes, with $T$ not at the point of intersection | B1 | Accept labels with delineators |
**NB: 2 separate diagrams scores max B1B0B1**
**Total: (3)**
---
## Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}T \cdot 3T$, $\quad \frac{(T+T-3.5)}{2} \cdot 14$ | M1 | Find distance for A or B in terms of $T$ only. Correct area formulae: must see $\frac{1}{2}$ in area formula and be adding in trapezium |
| One distance correct | A1 | |
| Both distances correct | A1 | |
| $\frac{1}{2}T \cdot 3T = \frac{(T+T-3.5)}{2} \cdot 14$ | M1 | Equate distances and simplify to a 3 term quadratic in $T$ in the form $aT^2 + bT + c = 0$ |
| $\frac{1}{2}T \cdot 3T = \frac{1}{2} \times 4 \times 3.5^2 + 14(T-3.5)$ | | |
| $3T^2 - 28T + 49 = 0$ | A1 | Correct quadratic |
| $(3T-7)(T-7) = 0$ | M1 | Solve 3 term quadratic for $T$ |
| $T = \frac{7}{3}$ or $7$ | A1 | Correct solution(s) - can be implied if only ever see $T=7$ from correct work |
| but $T > 3.5$, $\quad T = 7$ | A1 | |
**Total: (8)**
---
## Question 6(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 73.5 m | B1 | From correct work only. B0 if extra answers |
**Total: (1)**
---
## Question 6(e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Graph showing $A$ at velocity 4 (horizontal line) | B1 | Condone missing 4 |
| Graph showing $B$ (must go beyond 3.5, condone no 3) | B1 | Condone graph going beyond $T=7$ |
| Graph for $A$ going beyond $T=7$, dashed line shown | B1 | Condone graph going beyond $T=7$. B0 if see a solid vertical line |
**Condone separate diagrams.**
**Alternative for (c) with sketch showing $B$ at velocity $3T$:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Graphs for $A$ and $B$ | B1, B1, B0 | B1B1B0 on graph then max 5/8 for (c) if they do not solve for $T$ in the question |
| $\frac{1}{2} \times 3 \times (T+3.5)^2 = \frac{1}{2} \times 4 \times 3.5^2 + 14T$ | M1 | Use diagram to find area |
| One distance correct | A1 | |
| Both distances correct | A1 | |
| $12T^2 - 28T - 49 = 0$ | M1 | Simplify to a 3 term quadratic in $T$ |
| Correct quadratic | A1 | |
| $(2T-7)(6T+7) = 0$ | M1 | Complete method to solve for $T$ |
| $T = \frac{7}{2}$ or $\frac{-7}{6}$ | A1 | Correct solution(s) - implied if only see Total $= 7$ |
| Total time $= 7$ | A1 | |
**Total: (8)**
**Overall Total: (17 marks)**
---6.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{6ab8838f-d6f8-4761-8def-1022d97d4e82-16_264_997_269_461}
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\caption{Figure 2}
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Two cars, $A$ and $B$, move on parallel straight horizontal tracks. Initially $A$ and $B$ are both at rest with $A$ at the point $P$ and $B$ at the point $Q$, as shown in Figure 2. At time $t = 0$ seconds, $A$ starts to move with constant acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ for 3.5 s , reaching a speed of $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Car $A$ then moves with constant speed $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$.
Car $B$ also starts to move at time $t = 0$ seconds, in the same direction as car $A$. Car $B$ moves with a constant acceleration of $3 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. At time $t = T$ seconds, $B$ overtakes $A$. At this instant $A$ is moving with constant speed.
\item On a diagram, sketch, on the same axes, a speed-time graph for the motion of $A$ for the interval $0 \leqslant t \leqslant T$ and a speed-time graph for the motion of $B$ for the interval $0 \leqslant t \leqslant T$.
\item Find the value of $T$.
\item Find the distance of car $B$ from the point $Q$ when $B$ overtakes $A$.
\item On a new diagram, sketch, on the same axes, an acceleration-time graph for the motion of $A$ for the interval $0 \leqslant t \leqslant T$ and an acceleration-time graph for the motion of $B$ for the interval $0 \leqslant t \leqslant T$.
$\_\_\_\_$ VAYV SIHI NI JIIIM ION OC\\
VJYV SIHI NI JIIIM ION OC\\
VJYV SIHI NI JLIYM ION OC
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\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2018 Q6 [17]}}