| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding constants from motion conditions |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring basic differentiation and integration of polynomials, solving simultaneous equations to find constants, and applying initial conditions. All steps are routine M1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([q + r = 4\) and \(2q + 4r = 4]\) | M1 | Use \(v = 4\) at \(t = 1\) and \(t = 2\) |
| \(q = 6\) and \(r = -2\) so \(v = 6t - 2t^2\) | A1 | |
| \(a = 6 - 4t\) | M1 | Differentiation used for \(a\) |
| At \(t = 0.5\), \(a = 4\) | A1 | AG |
| Total: | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(v = 6t - 2t^2 = 0\) | M1 | Set \(v = 0\) and solve for \(t\) |
| \(t = 0\) and \(t = 3\) | A1 | |
| Total: | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| *EITHER:* \(s = \int(6t - 2t^2)\, dt\) | (M1) | Attempt to integrate \(v\) to find \(s\) |
| \(s = 3t^2 - \frac{2}{3}t^3 + C\) | A1 | |
| \(0 = 3 \times 3^2 - \frac{2}{3} \times 3^3 + C\) | M1 | Use \(s = 0\) when \(t = 3\) to find \(C\) |
| \(C = -9\) so distance \(= 9\) m | A1 | Valid argument |
| *OR:* \(s = \int_0^3 (6t - 2t^2)\, dt\) | (M1) | Attempt integration with limits |
| \(\left[3t^2 - \frac{2}{3}t^3\right]_0^3\) | A1 | Correct integration and correct limits but no evaluation |
| \([27 - 18 = 9]\) | M1 | Evaluation of integral between limits |
| Distance from \(O\) at \(t = 0\) is 9 m | A1 | With explanation |
| Total: | 4 |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[q + r = 4$ and $2q + 4r = 4]$ | M1 | Use $v = 4$ at $t = 1$ and $t = 2$ |
| $q = 6$ and $r = -2$ so $v = 6t - 2t^2$ | A1 | |
| $a = 6 - 4t$ | M1 | Differentiation used for $a$ |
| At $t = 0.5$, $a = 4$ | A1 | AG |
| **Total:** | **4** | |
---
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $v = 6t - 2t^2 = 0$ | M1 | Set $v = 0$ and solve for $t$ |
| $t = 0$ and $t = 3$ | A1 | |
| **Total:** | **2** | |
---
## Question 6(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| *EITHER:* $s = \int(6t - 2t^2)\, dt$ | (M1) | Attempt to integrate $v$ to find $s$ |
| $s = 3t^2 - \frac{2}{3}t^3 + C$ | A1 | |
| $0 = 3 \times 3^2 - \frac{2}{3} \times 3^3 + C$ | M1 | Use $s = 0$ when $t = 3$ to find $C$ |
| $C = -9$ so distance $= 9$ m | A1 | Valid argument |
| *OR:* $s = \int_0^3 (6t - 2t^2)\, dt$ | (M1) | Attempt integration with limits |
| $\left[3t^2 - \frac{2}{3}t^3\right]_0^3$ | A1 | Correct integration and correct limits but no evaluation |
| $[27 - 18 = 9]$ | M1 | Evaluation of integral between limits |
| Distance from $O$ at $t = 0$ is 9 m | A1 | With explanation |
| **Total:** | **4** | |
---6 A particle $P$ moves in a straight line passing through a point $O$. At time $t \mathrm {~s}$, the velocity of $P , v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, is given by $v = q t + r t ^ { 2 }$, where $q$ and $r$ are constants. The particle has velocity $4 \mathrm {~ms} ^ { - 1 }$ when $t = 1$ and when $t = 2$.\\
(i) Show that, when $t = 0.5$, the acceleration of $P$ is $4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
............................................................................................................................... .\\
(ii) Find the values of $t$ when $P$ is at instantaneous rest.\\
(iii) The particle is at $O$ when $t = 3$. Find the distance of $P$ from $O$ when $t = 0$.\\
\hfill \mbox{\textit{CAIE M1 2017 Q6 [10]}}