| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Session | Specimen |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity-time graph |
| Difficulty | Easy -1.2 This is a straightforward application of the area-under-graph interpretation of distance, requiring students to calculate the area of a trapezium (or split it into a rectangle and triangle). The question explicitly guides students to use the graph method and avoid SUVAT formulae, making it more accessible than typical mechanics problems. It's essentially testing understanding of a fundamental concept rather than problem-solving ability. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| Using distance \(=\) total area under graph (e.g. area of rectangle \(+\) triangle or trapezium or rectangle \(-\) triangle) | M1 | For use of distance \(=\) total area to give an equation in \(D\), \(U\), \(T\) and one other variable |
| e.g. \(D = UT + \frac{1}{2}Th\), where \(h\) is height of triangle | A1 | For a correct equation |
| Using gradient \(=\) acceleration to substitute \(h = aT\) | M1 | For using gradient \(= a\) to eliminate the other variable to give an equation in \(D\), \(U\), \(T\) and \(a\) only |
| \(D = UT + \frac{1}{2}aT^2\) | A1* | For a correct given answer |
## Question 6:
Using distance $=$ total area under graph (e.g. area of rectangle $+$ triangle **or** trapezium **or** rectangle $-$ triangle) | M1 | For use of distance $=$ total area to give an equation in $D$, $U$, $T$ and one other variable
e.g. $D = UT + \frac{1}{2}Th$, where $h$ is height of triangle | A1 | For a correct equation
Using gradient $=$ acceleration to substitute $h = aT$ | M1 | For using gradient $= a$ to eliminate the other variable to give an equation in $D$, $U$, $T$ and $a$ only
$D = UT + \frac{1}{2}aT^2$ | A1* | For a correct given answer6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8f3dbcb4-3260-4493-a230-12577b4ed691-12_520_1072_616_388}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A car moves along a straight horizontal road. At time $t = 0$, the velocity of the car is $U \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The car then accelerates with constant acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ for $T$ seconds. The car travels a distance $D$ metres during these $T$ seconds.
Figure 1 shows the velocity-time graph for the motion of the car for $0 \leqslant t \leqslant T$.\\
Using the graph, show that $D = U T + 1 / 2 a T ^ { 2 }$.\\
(No credit will be given for answers which use any of the kinematics (suvat) formulae listed under Mechanics in the AS Mathematics section of the formulae booklet.)
\hfill \mbox{\textit{Edexcel AS Paper 2 Q6 [4]}}