| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Standard +0.3 This is a standard variable acceleration problem requiring integration of a polynomial acceleration function with given boundary conditions. While it involves multiple integrations and finding a maximum using calculus, the techniques are routine for M1 level: integrate twice to get velocity and displacement, apply boundary conditions to find constants, and differentiate to find maximum speed. The algebra is straightforward and no novel insight is required. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| M1 | For using \(v(t) = \int a\, dt\) | |
| \(v = \frac{1}{160}t^3 - \frac{1}{3200}t^4 \quad (+C_1)\) | A1 | |
| \([0 = 8000/160 - 160000/3200 + C_1 \rightarrow C_1 = 0]\) | M1 | For using \(v(20) = 0\) |
| Initial speed is zero | A1 [4] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([t^2/800(15-t) = 0]\) | M1 | For solving \(a = 0\) |
| \(v_{\max} = v(15) = 5.27\ \text{ms}^{-1}\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| M1 | For using \(s(t) = \int v\, dt\) | |
| \(s = \frac{1}{640}t^4 - \frac{1}{16000}t^5 \quad (+C_2)\) | A1ft | |
| \([250 - 200]\) | M1 | For using limits 0 and 20 (or equivalent) |
| Distance AB is \(50\ \text{m}\) | A1 [4] |
## Question 7:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| | M1 | For using $v(t) = \int a\, dt$ |
| $v = \frac{1}{160}t^3 - \frac{1}{3200}t^4 \quad (+C_1)$ | A1 | |
| $[0 = 8000/160 - 160000/3200 + C_1 \rightarrow C_1 = 0]$ | M1 | For using $v(20) = 0$ |
| Initial speed is zero | A1 **[4]** | AG |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[t^2/800(15-t) = 0]$ | M1 | For solving $a = 0$ |
| $v_{\max} = v(15) = 5.27\ \text{ms}^{-1}$ | A1 **[2]** | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| | M1 | For using $s(t) = \int v\, dt$ |
| $s = \frac{1}{640}t^4 - \frac{1}{16000}t^5 \quad (+C_2)$ | A1ft | |
| $[250 - 200]$ | M1 | For using limits 0 and 20 (or equivalent) |
| Distance AB is $50\ \text{m}$ | A1 **[4]** | |7 A particle travels in a straight line from $A$ to $B$ in 20 s . Its acceleration $t$ seconds after leaving $A$ is $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, where $a = \frac { 3 } { 160 } t ^ { 2 } - \frac { 1 } { 800 } t ^ { 3 }$. It is given that the particle comes to rest at $B$.\\
(i) Show that the initial speed of the particle is zero.\\
(ii) Find the maximum speed of the particle.\\
(iii) Find the distance $A B$.
\hfill \mbox{\textit{CAIE M1 2011 Q7 [10]}}