| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Standard +0.3 This is a standard M1 variable acceleration question requiring differentiation to find maximum velocity (confirming k), integration for displacement, and evaluation at specific times. While it has multiple parts and requires careful algebraic manipulation, all techniques are routine for this level—finding stationary points, integrating polynomials, and substituting values. The multi-part structure and need to track when the particle returns to O elevates it slightly above average, but no novel problem-solving insight is required. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(dv/dt = k(120t - 3t^2)\) | B1 | |
| \([v(40) = k(60 \times 40^2 - 40^3) = 6.4]\) | M1 | For finding \(v_{\text{max}}\) as the value of \(v\) when \(dv/dt = 0\) and \(t \neq 0\) and equating with 6.4 |
| \(k = 0.0002\) | A1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = 60\) at \(A\) | B1 | |
| M1 | For integrating \(v(t)\) to find \(s(t)\) | |
| \(s(t) = 0.0002(20t^3 - t^4/4) \quad (+C)\) | A1 | |
| \([OA = 0.0002 \times (20 \times 60^3 - 60^4/4)]\) | M1 | For using limits 0 to 60 or evaluating \(s(t)\) when \(t = 60\) with \(C = 0\) (which may be implied by its absence) |
| Distance is 216 m | A1 | Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([dv/dt = 0.0002(120 \times 60 - 3 \times 60^2)]\) | M1 | For evaluating \(dv/dt\) when \(t = 60\) |
| Magnitude of acceleration is \(0.72 \text{ ms}^{-2}\) | A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(20t^3 - 0.25t^4 = 0\), \(v = 0.0002(60 \times 80^2 - 80^3)\) | M1 | For attempting to solve \(s(t) = 0\) for non-zero \(t\) and substituting into \(v(t)\) |
| Speed is \(25.6 \text{ ms}^{-1}\) | A1 | Total: 2 |
# Question 7:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $dv/dt = k(120t - 3t^2)$ | B1 | |
| $[v(40) = k(60 \times 40^2 - 40^3) = 6.4]$ | M1 | For finding $v_{\text{max}}$ as the value of $v$ when $dv/dt = 0$ and $t \neq 0$ and equating with 6.4 |
| $k = 0.0002$ | A1 | **Total: 3** | AG |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = 60$ at $A$ | B1 | |
| | M1 | For integrating $v(t)$ to find $s(t)$ |
| $s(t) = 0.0002(20t^3 - t^4/4) \quad (+C)$ | A1 | |
| $[OA = 0.0002 \times (20 \times 60^3 - 60^4/4)]$ | M1 | For using limits 0 to 60 or evaluating $s(t)$ when $t = 60$ with $C = 0$ (which may be implied by its absence) |
| Distance is 216 m | A1 | **Total: 5** |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[dv/dt = 0.0002(120 \times 60 - 3 \times 60^2)]$ | M1 | For evaluating $dv/dt$ when $t = 60$ |
| Magnitude of acceleration is $0.72 \text{ ms}^{-2}$ | A1 | **Total: 2** | Accept $a = -0.72 \text{ ms}^{-2}$ |
## Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $20t^3 - 0.25t^4 = 0$, $v = 0.0002(60 \times 80^2 - 80^3)$ | M1 | For attempting to solve $s(t) = 0$ for non-zero $t$ and substituting into $v(t)$ |
| Speed is $25.6 \text{ ms}^{-1}$ | A1 | **Total: 2** |7 A particle $P$ starts to move from a point $O$ and travels in a straight line. The velocity of $P$ is $k \left( 60 t ^ { 2 } - t ^ { 3 } \right) \mathrm { ms } ^ { - 1 }$ at time $t \mathrm {~s}$ after leaving $O$, where $k$ is a constant. The maximum velocity of $P$ is $6.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that $k = 0.0002$.\\
$P$ comes to instantaneous rest at a point $A$ on the line. Find\\
(ii) the distance $O A$,\\
(iii) the magnitude of the acceleration of $P$ at $A$,\\
(iv) the speed of $P$ when it subsequently passes through $O$.
\hfill \mbox{\textit{CAIE M1 2012 Q7 [12]}}