| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity-time graph |
| Difficulty | Moderate -0.5 This is a straightforward M1 question requiring students to recognize that displacement equals area under a velocity-time graph, then solve a simple trapezium area equation for V. It involves only one key concept and basic algebraic manipulation, making it slightly easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| When \(t = 2\), velocity is \(6 + 4 \times 2 = 14\) | M1 | Recognising that areas under graph represent changes in velocity in (i) or (ii) or equivalent suvat. |
| A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Require velocity of \(-6\) so must increase by \(-20\) | M1 | FT \(\pm(6 +\) their \(14)\) used in any attempt at area/suvat |
| \(-8 \times (t - 2) = -20\) so \(t = 4.5\) | F1 | FT their 14. [Award SC2 for \(4.5\) WW and SC1 for \(2.5\) WW] |
# Question 4:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| When $t = 2$, velocity is $6 + 4 \times 2 = 14$ | M1 | Recognising that areas under graph represent changes in velocity in (i) or (ii) or equivalent suvat. |
| | A1 | |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Require velocity of $-6$ so must increase by $-20$ | M1 | FT $\pm(6 +$ **their** $14)$ used in any attempt at area/suvat |
| $-8 \times (t - 2) = -20$ so $t = 4.5$ | F1 | FT **their** 14. [Award SC2 for $4.5$ WW and SC1 for $2.5$ WW] |4 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]{bdbebc7f-0cb1-4203-8058-7614ba291508-3_656_1344_401_399}
\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 Q4 [4]}}