| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Standard +0.3 This is a straightforward calculus-based mechanics question requiring standard techniques: finding when v=0 by solving a cubic equation (which factors easily), integrating velocity to find distance, and differentiating velocity to find maximum (setting dv/dt=0). All steps are routine applications of A-level methods with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using \(s = \int v\, dt\) | M1 | For using \(s = \int v\, dt\) |
| \(s = 0.027(10t^3/3 - t^4/4)\) \((+C)\) | A1 | |
| \(s = 0.027[10\,000/3 - 10000/4]\) | DM1 | For finding the value of \(t\) at A and using limits or equivalent |
| Distance is \(22.5\) m | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([0.027(20t - 3t^2) = 0 \Rightarrow t = 20/3]\) | M1 | For using \(dv/dt = 0\) |
| \(v_{\max} = 0.027(4000/9 - 8000/27)\) | A1ft | ft incorrect \(t\) in \(0.027(10t^2 - t^3)\) |
| Maximum speed is \(4\ \text{ms}^{-1}\) | A1 [3] |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $s = \int v\, dt$ | M1 | For using $s = \int v\, dt$ |
| $s = 0.027(10t^3/3 - t^4/4)$ $(+C)$ | A1 | |
| $s = 0.027[10\,000/3 - 10000/4]$ | DM1 | For finding the value of $t$ at A and using limits or equivalent |
| Distance is $22.5$ m | A1 [4] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[0.027(20t - 3t^2) = 0 \Rightarrow t = 20/3]$ | M1 | For using $dv/dt = 0$ |
| $v_{\max} = 0.027(4000/9 - 8000/27)$ | A1ft | ft incorrect $t$ in $0.027(10t^2 - t^3)$ |
| Maximum speed is $4\ \text{ms}^{-1}$ | A1 [3] | |
---3 A particle $P$ travels from a point $O$ along a straight line and comes to instantaneous rest at a point $A$. The velocity of $P$ at time $t \mathrm {~s}$ after leaving $O$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where $v = 0.027 \left( 10 t ^ { 2 } - t ^ { 3 } \right)$. Find\\
(i) the distance $O A$,\\
(ii) the maximum velocity of $P$ while moving from $O$ to $A$.
\hfill \mbox{\textit{CAIE M1 2012 Q3 [7]}}