| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity-time graph |
| Difficulty | Moderate -0.8 This is a straightforward application of the area-under-graph principle for velocity-time graphs. Students need only recognize that displacement equals area, calculate the trapezoid area using the given formula, and solve a simple linear equation for V. It requires basic recall and one-step algebraic manipulation, making it easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area of trapezium (or triangles) used | M1 | Must be attempt at area under graph |
| \(\frac{1}{2}(40+80)V + \frac{1}{2}(20)V = 1400\) or equivalent | A1 | Correct expression |
| \(60V + 10V = 1400\) | A1 | Simplification |
| \(V = 20\) | A1 |
# Question 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area of trapezium (or triangles) used | M1 | Must be attempt at area under graph |
| $\frac{1}{2}(40+80)V + \frac{1}{2}(20)V = 1400$ or equivalent | A1 | Correct expression |
| $60V + 10V = 1400$ | A1 | Simplification |
| $V = 20$ | A1 | |1 Fig. 1 is the velocity-time graph for the motion of a body. The velocity of the body is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t$ seconds.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{52d6c914-b204-4587-a82e-fbab6693fcf8-2_668_1360_461_354}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
The displacement of the body from $t = 0$ to $t = 100$ is 1400 m . Find the value of $V$.
\hfill \mbox{\textit{OCR MEI M1 2007 Q1 [4]}}