| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Displacement-time graph interpretation or sketching |
| Difficulty | Moderate -0.8 This is a straightforward kinematics problem using constant acceleration equations and area under velocity-time graph. Part (i) requires basic SUVAT application (distance = ½×6×12 + 10×12 = 156m) and sketching a standard displacement-time curve. Part (ii) uses the remaining distance (44m) with trapezium area formula to find V=10. All steps are routine applications of standard mechanics formulas with no problem-solving insight required, making it easier than average but not trivial due to the multi-step nature. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks |
|---|---|
| 5(i) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | Use constant acceleration equations or |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | Use constant acceleration equations or |
| Answer | Marks |
|---|---|
| 36 + 10 × 12 = 156 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| starting at (0 , 0) ending at (6 , 36) | B1 | Co-ordinates refer to (t,s) in a |
| Answer | Marks |
|---|---|
| (6 , 36) ends at (16 , 156) | B1 |
| Answer | Marks |
|---|---|
| (16 , 156) to (20 , 200) | B1 |
| Answer | Marks |
|---|---|
| 5(ii) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | Use relevant constant acceleration |
| Answer | Marks |
|---|---|
| V = 10 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
--- 5(i) ---
5(i) | 1
[s = (0 + 12) × 6]
1
2 | M1 | Use constant acceleration equations or
find area in (t,v) graph to find the distance
s travelled in the first 6 seconds
1
[s = 10 × 12]
2 | M1 | Use constant acceleration equations or
find area in (t,v) graph to find s the
2
distance travelled between 6s and 16s
Distance for first 16s is
36 + 10 × 12 = 156 | A1
Curve concave up for 0 < t < 6
starting at (0 , 0) ending at (6 , 36) | B1 | Co-ordinates refer to (t,s) in a
displacement-time graph
Line, positive gradient, 6 < t < 16 starts at
(6 , 36) ends at (16 , 156) | B1
Curve concave down, 16 < t < 20 from
(16 , 156) to (20 , 200) | B1
6
--- 5(ii) ---
5(ii) | 1
[44 = (12 + V) × 4]
2 | M1 | Use relevant constant acceleration
equations or the area property of a v–t
graph
V = 10 | A1
2
Question | Answer | Marks | GuidanceA sprinter runs a race of $200 \text{ m}$. His total time for running the race is $20 \text{ s}$. He starts from rest and accelerates uniformly for $6 \text{ s}$, reaching a speed of $12 \text{ m s}^{-1}$. He maintains this speed for the next $10 \text{ s}$, before decelerating uniformly to cross the finishing line with speed $V \text{ m s}^{-1}$.
\begin{enumerate}[label=(\roman*)]
\item Find the distance travelled by the sprinter in the first $16 \text{ s}$ of the race. Hence sketch a displacement-time graph for the $20 \text{ s}$ of the sprinter's race. [6]
\item Find the value of $V$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2018 Q5 [8]}}