| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Multi-phase journey: find unknown speed or time |
| Difficulty | Standard +0.3 This is a standard three-stage SUVAT problem requiring students to set up equations from a velocity-time graph, use kinematic formulas, and solve a quadratic. While it involves multiple stages and algebraic manipulation, it follows a well-practiced template with clear structure and routine techniques, making it slightly easier than average. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| B1 | Sketch requires three straight line segments with +ve, zero and –ve slopes in order, which together with a segment of the t axis form a trapezium | |
| M1 | For using \(v = at\) for \(T_1\) or \(u = -at\) for \(T_3\) | |
| \(T_1 = V \div 0.3,\ T_3 = V\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([S = \frac{1}{2}T_1 V + T_2 V + \frac{1}{2}T_3 V]\) | M1 | For using the area property for the distance travelled |
| M1 | For substituting for \(T_1\), \(T_2\) and \(T_3\) in terms of \(V\) | |
| \(S = 552V - V\{0.5(T_1 + T_3)\} = 552V - 13V^2/6\) | A1 | |
| \(13V^2 - 3312V + 72000 = 0\) | B1 | AG |
| \(V = 24\) | B1 | 5 marks |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| | B1 | Sketch requires three straight line segments with +ve, zero and –ve slopes in order, which together with a segment of the t axis form a trapezium |
| | M1 | For using $v = at$ for $T_1$ or $u = -at$ for $T_3$ |
| $T_1 = V \div 0.3,\ T_3 = V$ | A1 | 3 marks |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[S = \frac{1}{2}T_1 V + T_2 V + \frac{1}{2}T_3 V]$ | M1 | For using the area property for the distance travelled |
| | M1 | For substituting for $T_1$, $T_2$ and $T_3$ in terms of $V$ |
| $S = 552V - V\{0.5(T_1 + T_3)\} = 552V - 13V^2/6$ | A1 | |
| $13V^2 - 3312V + 72000 = 0$ | B1 | AG |
| $V = 24$ | B1 | 5 marks |
---5 A car travels in a straight line from $A$ to $B$, a distance of 12 km , taking 552 seconds. The car starts from rest at $A$ and accelerates for $T _ { 1 } \mathrm {~s}$ at $0.3 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, reaching a speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The car then continues to move at $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ for $T _ { 2 } \mathrm {~s}$. It then decelerates for $T _ { 3 } \mathrm {~s}$ at $1 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, coming to rest at $B$.\\
(i) Sketch the velocity-time graph for the motion and express $T _ { 1 }$ and $T _ { 3 }$ in terms of $V$.\\
(ii) Express the total distance travelled in terms of $V$ and show that $13 V ^ { 2 } - 3312 V + 72000 = 0$. Hence find the value of $V$.
\hfill \mbox{\textit{CAIE M1 2013 Q5 [8]}}