| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Displacement expressions and comparison |
| Difficulty | Standard +0.3 This is a standard M1 velocity-time graph question requiring interpretation of areas under graphs, basic SUVAT calculations, and sketching displacement-time graphs. While multi-part with several steps, each component uses routine techniques (area = displacement, solving for T, comparing displacements) without requiring novel insight or complex problem-solving beyond typical textbook exercises. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For correct shading of composite figure consisting of 2 rectangles: 1st has boundaries \(t=0\) & \(t=20\), \(v=0\) and \(v=2.5\); 2nd has boundaries \(t=20\) & \(t=T\), \(v=0\) and \(v=4\) | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([50 + 4(T-20) = 70\) or \(4T - 30 = 70]\) | M1 | For attempt to find equation in T |
| \(T = 25\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([\text{Distance} = 70 + (4-2.5)20\) or \(50 + 4[(T-20)+20]-50]\) | M1 | For identifying and using area representing required distance |
| Distance between P and Q is \(100\ \text{m}\) | A1ft [2] | ft 4T |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For 2 straight line segments representing P, 1st with +ve slope and 2nd with steeper slope, \(t=20\) indicated appropriately | B1 | |
| For Q, 1st & 2nd segments parallel to P's and displaced to the right, \(t=25\) and \(t=45\) indicated appropriately | B1ft [2] | ft T and \(T+20\) |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For correct shading of composite figure consisting of 2 rectangles: 1st has boundaries $t=0$ & $t=20$, $v=0$ and $v=2.5$; 2nd has boundaries $t=20$ & $t=T$, $v=0$ and $v=4$ | B1 **[1]** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[50 + 4(T-20) = 70$ or $4T - 30 = 70]$ | M1 | For attempt to find equation in T |
| $T = 25$ | A1 **[2]** | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[\text{Distance} = 70 + (4-2.5)20$ or $50 + 4[(T-20)+20]-50]$ | M1 | For identifying and using area representing required distance |
| Distance between P and Q is $100\ \text{m}$ | A1ft **[2]** | ft 4T |
### Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For 2 straight line segments representing P, 1st with +ve slope and 2nd with steeper slope, $t=20$ indicated appropriately | B1 | |
| For Q, 1st & 2nd segments parallel to P's and displaced to the right, $t=25$ and $t=45$ indicated appropriately | B1ft **[2]** | ft T and $T+20$ |
---(i) Make a rough copy of the diagram and shade the region whose area represents the displacement of $P$ from $X$ at the instant when $Q$ starts.
It is given that $P$ has travelled 70 m at the instant when $Q$ starts.\\
(ii) Find the value of $T$.\\
(iii) Find the distance between $P$ and $Q$ when $Q$ 's speed reaches $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(iv) Sketch a single diagram showing the displacement-time graphs for both $P$ and $Q$, with values shown on the $t$-axis at which the speed of either particle changes.
\hfill \mbox{\textit{CAIE M1 2011 Q4 [7]}}