| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Velocity-time graph sketching |
| Difficulty | Easy -1.3 This is a straightforward SUVAT question requiring only direct application of basic definitions. Part (a) involves sketching three horizontal line segments on a velocity-time graph from given information. Parts (b) and (c) require simple arithmetic: calculating areas under the graph for distance and using signs for displacement. No problem-solving, algebraic manipulation, or conceptual insight needed—purely routine mechanics at an introductory level. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Graph from 3 horizontal line segments with velocity \(4\) for \(0 \leq t \leq 5\), then \(-3.5\) for \(13 \leq t \leq 20\) | B1 | Graph from 3 horizontal line segments. Correct velocities labelled. Any lines joining the horizontal lines should be vertical |
| Times \(t = 5, 13, 20\) or lengths of line segments \(5, 8, 7\) seen | B1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distance \(= (4 \times 5) + (7 \times 3.5)\) m | M1 | Finding the area of at least one region from their graph oe. May work directly from the information in the question without reference to their graph |
| \(= 44.5\) | A1 [2] | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Displacement \(= 20 - 24.5 = -4.5\) m | B1 [1] | Allow for \(-4.5\) m or for \(4.5\) m south. Do not allow \(-4.5\) m south |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Graph from 3 horizontal line segments with velocity $4$ for $0 \leq t \leq 5$, then $-3.5$ for $13 \leq t \leq 20$ | B1 | Graph from 3 horizontal line segments. Correct velocities labelled. Any lines joining the horizontal lines should be vertical |
| Times $t = 5, 13, 20$ or lengths of line segments $5, 8, 7$ seen | B1 [2] | |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distance $= (4 \times 5) + (7 \times 3.5)$ m | M1 | Finding the area of at least one region from their graph oe. May work directly from the information in the question without reference to their graph |
| $= 44.5$ | A1 [2] | cao |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Displacement $= 20 - 24.5 = -4.5$ m | B1 [1] | Allow for $-4.5$ m or for $4.5$ m south. Do not allow $-4.5$ m south |
---5 A child is running up and down a path. A simplified model of the child's motion is as follows:
\begin{itemize}
\item he first runs north for 5 s at $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$;
\item he then suddenly stops and waits for 8 s ;
\item finally he runs in the opposite direction for 7 s at $3.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Taking north to be the positive direction, sketch a velocity-time graph for this model of the child's motion.
\end{itemize}
Using this model,
\item calculate the total distance travelled by the child,
\item find his final displacement from his original position.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2020 Q5 [5]}}