| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a standard multi-part kinematics question requiring differentiation to find acceleration, then applying constant acceleration equations (suvat). While it has multiple parts and requires careful tracking of values between sections, each individual step uses routine A-level mechanics techniques with no novel problem-solving or geometric insight required. Slightly above average due to the multi-stage nature and bookkeeping involved. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = 0.5 - 0.02t\) | B1 | |
| \([0.5 - 0.02t = 0.1]\) | M1 | For solving \(\frac{dv}{dt} = 0.1\) |
| Time taken is 20 s | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(u = 0.5 \times 20 - 0.01 \times 20^2 (=\frac{v}{t})\) | B1ft | ft 0.5\(t_1\) − 0.01\(t_1^2\) |
| \([14 = 6 + 0.1t]\) | M1 | For using \(v = u + at\) |
| Time taken is 80 s | A1ft | [3] ft \(t = 10(14 - 0.5t_1 + 0.01t_1^2)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \([v^2 = 14^2 - 2 \times 0.3 \times 300]\) | M1 | For using \(v^2 = u^2 + 2as\) |
| Speed is 4 ms⁻¹ | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(s = 0.25t^2 - 0.01t^3/3 (+ C)\) | M1 | For using \(s = \int v\mathrm{d}t\) |
| \(AB = 0.25 \times 20^2 - 0.01 \times 20^3/3 (= 73.3)\) | A1 | |
| \(BC = \frac{1}{2}(6 + 14) \times 80 \text{ or } 6 \times 80 + \frac{1}{2} \times 0.1 \times 80^2\) | DM1 | For using limits 0 to 20 or equivalent |
| \((\text{or } (14^2 - 6^2)/(2 \times 0.1) (= 800)\) | ft 0.25\(t_1^2\) − 0.01\(t_1^3\)/3 | |
| Distance AD is 1170 m | A1 | [6] |
**(i)**
$a = 0.5 - 0.02t$ | B1 |
$[0.5 - 0.02t = 0.1]$ | M1 | For solving $\frac{dv}{dt} = 0.1$
Time taken is 20 s | A1 | [3]
**(ii)**
$u = 0.5 \times 20 - 0.01 \times 20^2 (=\frac{v}{t})$ | B1ft | ft 0.5$t_1$ − 0.01$t_1^2$
$[14 = 6 + 0.1t]$ | M1 | For using $v = u + at$
Time taken is 80 s | A1ft | [3] ft $t = 10(14 - 0.5t_1 + 0.01t_1^2)$
**(iii)**
$[v^2 = 14^2 - 2 \times 0.3 \times 300]$ | M1 | For using $v^2 = u^2 + 2as$
Speed is 4 ms⁻¹ | A1 | [2]
**(iv)**
$s = 0.25t^2 - 0.01t^3/3 (+ C)$ | M1 | For using $s = \int v\mathrm{d}t$
$AB = 0.25 \times 20^2 - 0.01 \times 20^3/3 (= 73.3)$ | A1 |
$BC = \frac{1}{2}(6 + 14) \times 80 \text{ or } 6 \times 80 + \frac{1}{2} \times 0.1 \times 80^2$ | DM1 | For using limits 0 to 20 or equivalent
$(\text{or } (14^2 - 6^2)/(2 \times 0.1) (= 800)$ | | ft 0.25$t_1^2$ − 0.01$t_1^3$/3
Distance AD is 1170 m | A1 | [6]A particle $P$ travels in a straight line from $A$ to $D$, passing through the points $B$ and $C$. For the section $AB$ the velocity of the particle is $(0.5t - 0.01t^2)$ m s$^{-1}$, where $t$ s is the time after leaving $A$.
\begin{enumerate}[label=(\roman*)]
\item Given that the acceleration of $P$ at $B$ is 0.1 m s$^{-2}$, find the time taken for $P$ to travel from $A$ to $B$. [3]
\end{enumerate}
The acceleration of $P$ from $B$ to $C$ is constant and equal to 0.1 m s$^{-2}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Given that $P$ reaches $C$ with speed 14 m s$^{-1}$, find the time taken for $P$ to travel from $B$ to $C$. [3]
\end{enumerate}
$P$ travels with constant deceleration 0.3 m s$^{-2}$ from $C$ to $D$. Given that the distance $CD$ is 300 m, find
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item the speed with which $P$ reaches $D$, [2]
\item the distance $AD$. [6]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2009 Q7 [14]}}