| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity-time graph |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics question requiring basic interpretation of a velocity-time graph (area under curve for distance), simple arithmetic with speed-distance-time, and direct application of F=ma. All parts are standard textbook exercises with no problem-solving insight needed, making it easier than average A-level questions. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(s_1=\frac{1}{2}\times15\times20=150\) | M1 | Finding length of first stage |
| \(s_2=\frac{1}{2}\times15\times80=600\) | M1 | Finding length of second stage |
| A1 | Both distances correct | |
| \(s=600+150=750\text{ m}\) | A1 | Correct total distance |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(t=\frac{750}{15}=50\text{ s}\) | B1ft | Correct time or their distance correctly divided by 15 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\text{Delay}=120-50=70\text{ s}\) | B1ft | Correct time or their previous time correctly subtracted from 120 to give a positive answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(a=\frac{15}{80}=\frac{3}{16}=0.1875\text{ ms}^{-2}\) | M1 | Finding acceleration |
| A1 | Correct acceleration | |
| \(F=500000\times0.1875=93800\text{ N}\) (to 3sf) | M1 | Use of \(F=ma\) |
| A1 | Correct force |
## Question 2:
**Part (a)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $s_1=\frac{1}{2}\times15\times20=150$ | M1 | Finding length of first stage |
| $s_2=\frac{1}{2}\times15\times80=600$ | M1 | Finding length of second stage |
| | A1 | Both distances correct |
| $s=600+150=750\text{ m}$ | A1 | Correct total distance |
**Part (b)(i)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $t=\frac{750}{15}=50\text{ s}$ | B1ft | Correct time or their distance correctly divided by 15 |
**Part (b)(ii)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $\text{Delay}=120-50=70\text{ s}$ | B1ft | Correct time or their previous time correctly subtracted from 120 to give a positive answer |
**Part (c)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $a=\frac{15}{80}=\frac{3}{16}=0.1875\text{ ms}^{-2}$ | M1 | Finding acceleration |
| | A1 | Correct acceleration |
| $F=500000\times0.1875=93800\text{ N}$ (to 3sf) | M1 | Use of $F=ma$ |
| | A1 | Correct force |
---2 A train travels along a straight horizontal track between two points $A$ and $B$.\\
Initially the train is at $A$ and moving at $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Due to a problem, the train has to slow down and stop. At time $t = 40$ seconds it begins to move again. At time $t = 120$ seconds the train is at $B$ and moving at $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ again.
The graph below shows how the velocity of the train varies as it moves from $A$ to $B$.\\
\includegraphics[max width=\textwidth, alt={}, center]{7e0585ea-062a-487c-8e39-37a4ed414ff8-2_408_1086_1505_434}
\begin{enumerate}[label=(\alph*)]
\item Use the graph to find the total distance between the points $A$ and $B$.
\item The train should have travelled between $A$ and $B$ at a constant velocity of $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\roman*)]
\item Calculate the time that the train would take to travel between $A$ and $B$ at a speed of $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Calculate the time by which the train was delayed.
\end{enumerate}\item The train has mass 500 tonnes. Find the resultant force acting on the train when $40 < t < 120$.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2005 Q2 [10]}}