| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Two-particle meeting or overtaking |
| Difficulty | Standard +0.3 This is a standard kinematics problem requiring systematic application of SUVAT equations across three phases of motion, followed by a straightforward overtaking calculation. While it involves multiple steps and careful bookkeeping, it requires no novel insight—just methodical application of well-practiced techniques typical of M1 coursework. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| 7(i) | 36 = 0 + 0.5 × 0.5t2 | |
| t = 12 | 1 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| v = 6 | 1 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| = 210 – 36 – 150 = 24 | 1 | B1 |
| 24 = (6 + 0) / 2 × t | 1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Total Time = 12 + 25 + 8 = 45 s | 1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7(ii) | Distance travelled by cyclist | |
| = 36 + 6(t – 12) | 1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| = 0.5 × 4 × (t – 24)2 | 1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [t2 – 51t + 594 = 0] | 1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| t = 33 or t = 18 | 1 | A1 |
| Time = 33 s | 1 | B1 |
Question 7:
--- 7(i) ---
7(i) | 36 = 0 + 0.5 × 0.5t2
t = 12 | 1 | B1
v2 = 0 + 2 × 0.5 × 36
v = 6 | 1 | B1
s = 6 × 25
remaining distance
= 210 – 36 – 150 = 24 | 1 | B1
24 = (6 + 0) / 2 × t | 1 | M1 | Using s = (u + v)t / 2
t = 8
Total Time = 12 + 25 + 8 = 45 s | 1 | A1
5
Marks
--- 7(ii) ---
7(ii) | Distance travelled by cyclist
= 36 + 6(t – 12) | 1 | M1 | For attempting distance travelled by cyclist for t > 12
Distance travelled by car
= 0.5 × 4 × (t – 24)2 | 1 | M1 | For attempting distance travelled by car
2t2 – 96t + 1152
= 36 + 6t – 72
[t2 – 51t + 594 = 0] | 1 | M1 | Equating expressions and attempting to solve a three
term quadratic equation
t = 33 or t = 18 | 1 | A1
Time = 33 s | 1 | B1 | Choosing the correct solution
5A cyclist starts from rest at point $A$ and moves in a straight line with acceleration 0.5 m s$^{-2}$ for a distance of 36 m. The cyclist then travels at constant speed for 25 s before slowing down, with constant deceleration, to come to rest at point $B$. The distance $AB$ is 210 m.
\begin{enumerate}[label=(\roman*)]
\item Find the total time that the cyclist takes to travel from $A$ to $B$. [5]
\item Find the time that it takes from when the cyclist starts until the car overtakes her. [5]
\end{enumerate}
24 s after the cyclist leaves point $A$, a car starts from rest from point $A$, with constant acceleration 4 m s$^{-2}$ towards $B$. It is given that the car overtakes the cyclist while the cyclist is moving with constant speed.
\hfill \mbox{\textit{CAIE M1 Q7 [10]}}