| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Time to reach midpoint or specific position |
| Difficulty | Standard +0.3 This is a standard M1 SUVAT question with a trapezoid speed-time graph. Parts (a)-(c) involve routine application of area under graph equals distance, requiring only algebraic manipulation to find V and then time to midpoint. Part (d) adds a minor twist by asking when a specific speed occurs twice, but this is still straightforward substitution into SUVAT equations. Slightly easier than average due to the structured, predictable nature of the multi-part question with clear signposting. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Trapezium shape starting and finishing on \(t\)-axis | B1 | Not to scale; B0 if solid vertical lines included |
| Correct figures (60, 270, use of 30 with 240) and \(V\) marked | B1 | 270 can be implied by 3 correct delineators |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4500 = \frac{(270+180)}{2}V\) OR \(4500 = \frac{1}{2}60V + 180V + \frac{1}{2}30V\) | M1 A1 | Complete method for equation in \(V\) only; one trapezium or two triangles + rectangle etc.; M0 if single *suvat* used |
| \(V = 20\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{(T+T-60)}{2} \times 20 = 2250\) OR \(\frac{1}{2}60.20 + (T-60).20 = 2250\) | M1 A2 ft | Complete method in ONE variable; one trapezium or triangle + rectangle; allow 2.25 for M mark; M0 if single *suvat* used; ft on their 20; -1 each error |
| \(T = 142.5\) s | A1 | Accept 143 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T_1 = \frac{1}{4} \times 60\) | M1 | Complete method for equation in \(T_1\) only |
| \(= 15\) | A1 | Independent of \(V\) |
| \(T_2 = 270 - \left(\frac{1}{4} \times 30\right)\) OR \(240 + \left(\frac{3}{4} \times 30\right)\) | M1 A1 | Complete method for equation in \(T_2\) only |
| \(= 262.5\) | A1 | Independent of \(V\); accept 263. N.B. Accept \(T_1 = 262.5\) and \(T_2 = 15\) |
## Question 6:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Trapezium shape starting and finishing on $t$-axis | B1 | Not to scale; B0 if solid vertical lines included |
| Correct figures (60, 270, use of 30 with 240) and $V$ marked | B1 | 270 can be implied by 3 correct delineators |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4500 = \frac{(270+180)}{2}V$ OR $4500 = \frac{1}{2}60V + 180V + \frac{1}{2}30V$ | M1 A1 | Complete method for equation in $V$ only; one trapezium or two triangles + rectangle etc.; M0 if single *suvat* used |
| $V = 20$ | A1 | |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{(T+T-60)}{2} \times 20 = 2250$ OR $\frac{1}{2}60.20 + (T-60).20 = 2250$ | M1 A2 ft | Complete method in ONE variable; one trapezium or triangle + rectangle; allow 2.25 for M mark; M0 if single *suvat* used; ft on their 20; -1 each error |
| $T = 142.5$ s | A1 | Accept 143 |
### Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T_1 = \frac{1}{4} \times 60$ | M1 | Complete method for equation in $T_1$ only |
| $= 15$ | A1 | Independent of $V$ |
| $T_2 = 270 - \left(\frac{1}{4} \times 30\right)$ OR $240 + \left(\frac{3}{4} \times 30\right)$ | M1 A1 | Complete method for equation in $T_2$ only |
| $= 262.5$ | A1 | Independent of $V$; accept 263. N.B. Accept $T_1 = 262.5$ and $T_2 = 15$ |
---6. A train travels for a total of 270 s along a straight horizontal track between two stations $A$ and $B$. The train starts from rest at $A$ and moves with constant acceleration for 60 s until it reaches a speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The train then travels at this constant speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ before it moves with constant deceleration for 30 s , coming to rest at $B$.
\begin{enumerate}[label=(\alph*)]
\item Sketch below a speed-time graph for the journey of the train between the two stations $A$ and $B$.
Given that the distance between the two stations is 4.5 km ,
\item find the value of $V$,
\item find how long it takes the train to travel from station $A$ to the point that is exactly halfway between the two stations.
The train is travelling at speed $\frac { 1 } { 4 } V \mathrm {~ms} ^ { - 1 }$ at times $T _ { 1 }$ seconds and $T _ { 2 }$ seconds after leaving station $A$.
\item Find the value of $T _ { 1 }$ and the value of $T _ { 2 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2019 Q6 [14]}}