| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding when particle at rest |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring standard integration techniques. Part (i) involves finding when v=0, then integrating velocity to get displacement—routine calculus. Part (ii) requires double integration of acceleration with initial conditions, which is a standard textbook exercise. The calculations are direct with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6t - 0.3t^2 = 0 \rightarrow t = 20\) (or \(0\)) | B1 | |
| \([s = 6t^2/2 - 0.3t^3/3\ (+C)]\) | M1 | For integrating \(v(t)\) to obtain \(s(t)\) |
| \([s = 6(20)^2/2 - 0.3(20)^3/3]\) | DM1 | For evaluating \(s(t)\) when \(v = 0\) |
| Distance OX is \(400\text{ m}\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([v = kt - 6t^2\ (+C)]\) | M1* | For integrating \(a(t)\) to obtain \(v(t)\) |
| \([s = kt^2/2 - 6t^3/3]\) | M1* | For integrating \(v(t)\) to obtain \(s(t)\) and for using \(s(0) = 0\) |
| \([400 = 0.5k \times 10^2 - 2 \times 10^3]\) | DM1 | For using \(t = 10\) and \(s = 400\) to form equation in \(k\) |
| \(k = 48\) | A1 | [4] |
# Question 5:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6t - 0.3t^2 = 0 \rightarrow t = 20$ (or $0$) | B1 | |
| $[s = 6t^2/2 - 0.3t^3/3\ (+C)]$ | M1 | For integrating $v(t)$ to obtain $s(t)$ |
| $[s = 6(20)^2/2 - 0.3(20)^3/3]$ | DM1 | For evaluating $s(t)$ when $v = 0$ |
| Distance OX is $400\text{ m}$ | A1 | [4] |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[v = kt - 6t^2\ (+C)]$ | M1* | For integrating $a(t)$ to obtain $v(t)$ |
| $[s = kt^2/2 - 6t^3/3]$ | M1* | For integrating $v(t)$ to obtain $s(t)$ and for using $s(0) = 0$ |
| $[400 = 0.5k \times 10^2 - 2 \times 10^3]$ | DM1 | For using $t = 10$ and $s = 400$ to form equation in $k$ |
| $k = 48$ | A1 | [4] |
---5 A particle $P$ starts from a fixed point $O$ and moves in a straight line. At time $t$ s after leaving $O$, the velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ of $P$ is given by $v = 6 t - 0.3 t ^ { 2 }$. The particle comes to instantaneous rest at point $X$.\\
(i) Find the distance $O X$.
A second particle $Q$ starts from rest from $O$, at the same instant as $P$, and also travels in a straight line. The acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ of $Q$ is given by $a = k - 12 t$, where $k$ is a constant. The displacement of $Q$ from $O$ is 400 m when $t = 10$.\\
(ii) Find the value of $k$.
\hfill \mbox{\textit{CAIE M1 2016 Q5 [8]}}