| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity-time graph |
| Difficulty | Easy -1.2 This is a straightforward velocity-time graph question requiring only basic interpretation skills: reading values directly from the graph and calculating areas of simple geometric shapes (triangles and rectangles). All parts are routine recall and application of the fundamental principle that distance = area under v-t graph, with no problem-solving or novel insight required. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| \(30\) seconds | B1 | Time from \(t=40\) to \(t=70\) |
| Answer | Marks | Guidance |
|---|---|---|
| Area of triangle = \(\frac{1}{2} \times 40 \times 20\) | M1 | Area under graph for first 40s |
| \(= 400\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Distance from \(t=70\) to \(t=120\): \(\frac{1}{2} \times 50 \times 20 = 500\) m | M1 | Area of triangle for acceleration phase |
| Total = \(400 + 0 + 500 = 900\) m | A1 |
| Answer | Marks |
|---|---|
| Average speed \(= \frac{900}{120}\) | M1 |
| \(= 7.5\) m s\(^{-1}\) | A1 |
| Answer | Marks |
|---|---|
| Distance at constant \(20\) m s\(^{-1}\): \(20 \times 120 = 2400\) m | M1 |
| Extra distance \(= 2400 - 900 = 1500\) m | A1 |
# Question 1:
**(a)**
$30$ seconds | B1 | Time from $t=40$ to $t=70$
**(b)**
Area of triangle = $\frac{1}{2} \times 40 \times 20$ | M1 | Area under graph for first 40s
$= 400$ m | A1 |
**(c)**
Distance from $t=70$ to $t=120$: $\frac{1}{2} \times 50 \times 20 = 500$ m | M1 | Area of triangle for acceleration phase
Total = $400 + 0 + 500 = 900$ m | A1 |
**(d)**
Average speed $= \frac{900}{120}$ | M1 |
$= 7.5$ m s$^{-1}$ | A1 |
**(e)**
Distance at constant $20$ m s$^{-1}$: $20 \times 120 = 2400$ m | M1 |
Extra distance $= 2400 - 900 = 1500$ m | A1 |
---1 A bus slows down as it approaches a bus stop. It stops at the bus stop and remains at rest for a short time as the passengers get on. It then accelerates away from the bus stop. The graph shows how the velocity of the bus varies.\\
\includegraphics[max width=\textwidth, alt={}, center]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-02_627_1296_657_402}
Assume that the bus travels in a straight line during the motion described by the graph.
\begin{enumerate}[label=(\alph*)]
\item State the length of time for which the bus is at rest.
\item Find the distance travelled by the bus in the first 40 seconds.
\item Find the total distance travelled by the bus in the 120 -second period.
\item Find the average speed of the bus in the 120 -second period.
\item If the bus had not stopped but had travelled at a constant $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ for the 120 -second period, how much further would it have travelled?
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-03_2484_1709_223_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2010 Q1 [9]}}