| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Displacement-time graph interpretation or sketching |
| Difficulty | Standard +0.3 This is a structured multi-part question requiring interpretation of a displacement-time graph and conversion to velocity-time, with straightforward SUVAT calculations. While it involves multiple steps and understanding of tangent gradients representing velocity, each part follows standard mechanics procedures without requiring novel insight or complex problem-solving beyond typical M1 level. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v\)-\(t\) graph with 3 straight line segments | M1 | For 3 straight line segments; \(v(t)\) positive for \(0 < t < 600\), continuous and single valued |
| End points at \(t=0\), \(t=600\) on \(t\)-axis | A1 | End points (\(t=0\), \(t=600\)) on \(t\)-axis |
| \(t = 80\), \(t = 560\), \(t = 600\) marked; +ve, zero, \(-\)ve gradients in order | A1 | \(+\)ve, zero and \(-\)ve gradients in order |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2} \times 80 \times v = 400\) | M1 | For using the idea that area of triangle represents distance, or using \(\frac{(0+v)}{2} = s \div t\) |
| Velocity is \(10\ \text{ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(D = \frac{1}{2}(600 + 480) \times 10\) | M1 | For using the idea that area of trapezium represents total distance |
| Total distance is \(5400\) m | A1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Acceleration is \(0.125\ \text{ms}^{-2}\) for \(0 < t < 80\) | M1, A1ft | For using the idea that gradient represents acceleration, or for using \(v = 0 + at\) |
## Question 5:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v$-$t$ graph with 3 straight line segments | M1 | For 3 straight line segments; $v(t)$ positive for $0 < t < 600$, continuous and single valued |
| End points at $t=0$, $t=600$ on $t$-axis | A1 | End points ($t=0$, $t=600$) on $t$-axis |
| $t = 80$, $t = 560$, $t = 600$ marked; +ve, zero, $-$ve gradients in order | A1 | $+$ve, zero and $-$ve gradients in order |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times 80 \times v = 400$ | M1 | For using the idea that area of triangle represents distance, or using $\frac{(0+v)}{2} = s \div t$ |
| Velocity is $10\ \text{ms}^{-1}$ | A1 | |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $D = \frac{1}{2}(600 + 480) \times 10$ | M1 | For using the idea that area of trapezium represents total distance |
| Total distance is $5400$ m | A1ft | |
### Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Acceleration is $0.125\ \text{ms}^{-2}$ for $0 < t < 80$ | M1, A1ft | For using the idea that gradient represents acceleration, or for using $v = 0 + at$ |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{2026cad4-8494-4139-ad21-d8a17ac2b955-3_917_1451_1059_347}
The diagram shows the displacement-time graph for a car's journey. The graph consists of two curved parts $A B$ and $C D$, and a straight line $B C$. The line $B C$ is a tangent to the curve $A B$ at $B$ and a tangent to the curve $C D$ at $C$. The gradient of the curves at $t = 0$ and $t = 600$ is zero, and the acceleration of the car is constant for $0 < t < 80$ and for $560 < t < 600$. The displacement of the car is 400 m when $t = 80$.\\
(i) Sketch the velocity-time graph for the journey.\\
(ii) Find the velocity at $t = 80$.\\
(iii) Find the total distance for the journey.\\
(iv) Find the acceleration of the car for $0 < t < 80$.
\hfill \mbox{\textit{CAIE M1 2005 Q5 [9]}}