| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Constant acceleration with algebraic unknowns |
| Difficulty | Moderate -0.8 This is a straightforward M1 SUVAT question with three parts that progressively add complexity but use standard formulas (v² = u² + 2as, v = u + at, s = ut + ½at²) in routine ways. Each part is clearly signposted with no conceptual difficulty or problem-solving insight required—just careful application of memorized equations. Easier than average A-level maths due to its mechanical, step-by-step nature. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v^2 = 25^2 - 2 \times 6 \times 48\) | M1 | Complete method to find \(v\), condone sign errors |
| \(v = 7\ (\text{m s}^{-1})\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{25-13}{6}\ (= 2)\) | M1 | Complete method to find time to reach \(13\ \text{ms}^{-1}\) |
| \(13^2 = 25^2 - 2 \times 6s\) OR \(25 \times 2 - \dfrac{1}{2} \times 6 \times 2^2\) OR \(\dfrac{(25+13)}{2} \times 2\) OR \(13 \times 2 - \dfrac{1}{2} \times (-6) \times 2^2\ (s=38)\) | M1 | Complete method to find distance travelled in reaching \(13\ \text{ms}^{-1}\), ft on their 2 if necessary |
| Total time \(= \dfrac{(48-38)}{13} + 2\) | DM1 | Dependent on previous two M marks, complete method to find total time, ft on their 2 and 38 |
| \(\dfrac{36}{13} = 2\dfrac{10}{13}\ \text{(s)}\ (2.76923...)\) | A1 | Correct answer. Allow 2.8 or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{25-13}{6}\ (= 2)\) (could be implied by 2.2) | M1 | Complete method to find time to reach \(13\ \text{ms}^{-1}\) once it starts decelerating |
| \((0.2 \times 25) + \left(25 \times 2 - \dfrac{1}{2} \times 6 \times 2^2\right)\ (5+38)\) | M1 | Complete method to find total distance travelled in reaching \(13\ \text{ms}^{-1}\), ft on their 2 |
| Total time \(= \dfrac{48 - [(0.2 \times 25) + 38]}{13} + 0.2 + 2\) | DM1 | Dependent on previous two M marks, ft on their 2 and 38 |
| \(\dfrac{168}{65} = 2\dfrac{38}{65}\ \text{(s)}\ (2.58461538...)\) | A1 | Correct answer. Allow 2.6 or better |
# Question 3:
## Part 3(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v^2 = 25^2 - 2 \times 6 \times 48$ | M1 | Complete method to find $v$, condone sign errors |
| $v = 7\ (\text{m s}^{-1})$ | A1 | cao |
## Part 3(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{25-13}{6}\ (= 2)$ | M1 | Complete method to find time to reach $13\ \text{ms}^{-1}$ |
| $13^2 = 25^2 - 2 \times 6s$ OR $25 \times 2 - \dfrac{1}{2} \times 6 \times 2^2$ OR $\dfrac{(25+13)}{2} \times 2$ OR $13 \times 2 - \dfrac{1}{2} \times (-6) \times 2^2\ (s=38)$ | M1 | Complete method to find distance travelled in reaching $13\ \text{ms}^{-1}$, ft on their 2 if necessary |
| Total time $= \dfrac{(48-38)}{13} + 2$ | DM1 | Dependent on previous two M marks, complete method to find total time, ft on their 2 and 38 |
| $\dfrac{36}{13} = 2\dfrac{10}{13}\ \text{(s)}\ (2.76923...)$ | A1 | Correct answer. Allow 2.8 or better |
## Part 3(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{25-13}{6}\ (= 2)$ (could be implied by 2.2) | M1 | Complete method to find time to reach $13\ \text{ms}^{-1}$ once it starts decelerating |
| $(0.2 \times 25) + \left(25 \times 2 - \dfrac{1}{2} \times 6 \times 2^2\right)\ (5+38)$ | M1 | Complete method to find total distance travelled in reaching $13\ \text{ms}^{-1}$, ft on their 2 |
| Total time $= \dfrac{48 - [(0.2 \times 25) + 38]}{13} + 0.2 + 2$ | DM1 | Dependent on previous two M marks, ft on their 2 and 38 |
| $\dfrac{168}{65} = 2\dfrac{38}{65}\ \text{(s)}\ (2.58461538...)$ | A1 | Correct answer. Allow 2.6 or better |
---3. A car is moving at a constant speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ along a straight horizontal road.
The car is modelled as a particle.\\
At time $t = 0$, the car is at the point $A$ and the driver sees a road sign 48 m ahead.\\
Let $t$ seconds be the time that elapses after the car passes $A$.\\
In a first model, the car is assumed to decelerate uniformly at $6 \mathrm {~ms} ^ { - 2 }$ from $A$ until the car reaches the road sign.
\begin{enumerate}[label=(\alph*)]
\item Use this first model to find the speed of the car as it reaches the sign.
The road sign indicates that the speed limit immediately after the sign is $13 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
In a second model, the car is assumed to decelerate uniformly at $6 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ from $A$ until it reaches a speed of $13 \mathrm {~ms} ^ { - 1 }$. The car then maintains this speed until it reaches the road sign.
\item Use this second model to find the value of $t$ at which the car reaches the sign.
In a third model, the car is assumed to move with constant speed $25 \mathrm {~ms} ^ { - 1 }$ from $A$ until time $t = 0.2$, the car then decelerates uniformly at $6 \mathrm {~ms} ^ { - 2 }$ until it reaches a speed of $13 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The car then maintains this speed until it reaches the road sign.
\item Use this third model to find the value of $t$ at which the car reaches the sign.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2021 Q3 [10]}}