| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding when particle at rest |
| Difficulty | Standard +0.3 This is a standard mechanics question on non-constant acceleration requiring differentiation of velocity for acceleration, solving a quadratic for rest positions, and integration for displacement. All techniques are routine A-level methods 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 |
| \(a = 12t - 30\) | M1 | For differentiating \(v\) to find \(a\) |
| \(t < 2.5\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(v = 0\) at \(t = 1\) and \(t = 4\) | B1 | Using \(v = 6(t-4)(t-1)\) |
| \(s = \int(6t^2 - 30t + 24)\,dt\) | M1 | For using integration to find \(s\) |
| \(= \frac{6}{3}t^3 - \frac{30}{2}t^2 + 24t\) | ||
| \(s = \left[2t^3 - 15t^2 + 24t\right]_1^4\) | M1 | For using limits |
| Distance \(= 27\) m | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2t^3 - 15t^2 + 24t = 0\) | M1 | State \(s = 0\) |
| \(2t^2 - 15t + 24 = 0\) | M1 | Reduce to a quadratic and attempt to solve |
| \(t = 2.31\) and \(t = 5.19\) | A1 | [3] |
## Question 6:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $a = 12t - 30$ | M1 | For differentiating $v$ to find $a$ |
| $t < 2.5$ | A1 | [2] |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $v = 0$ at $t = 1$ and $t = 4$ | B1 | Using $v = 6(t-4)(t-1)$ |
| $s = \int(6t^2 - 30t + 24)\,dt$ | M1 | For using integration to find $s$ |
| $= \frac{6}{3}t^3 - \frac{30}{2}t^2 + 24t$ | | |
| $s = \left[2t^3 - 15t^2 + 24t\right]_1^4$ | M1 | For using limits |
| Distance $= 27$ m | A1 | [4] |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $2t^3 - 15t^2 + 24t = 0$ | M1 | State $s = 0$ |
| $2t^2 - 15t + 24 = 0$ | M1 | Reduce to a quadratic and attempt to solve |
| $t = 2.31$ and $t = 5.19$ | A1 | [3] |6 A particle $P$ moves in a straight line. It starts at a point $O$ on the line and at time $t$ s after leaving $O$ it has a velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where $v = 6 t ^ { 2 } - 30 t + 24$.\\
(i) Find the set of values of $t$ for which the acceleration of the particle is negative.\\
(ii) Find the distance between the two positions at which $P$ is at instantaneous rest.\\
(iii) Find the two positive values of $t$ at which $P$ passes through $O$.
\hfill \mbox{\textit{CAIE M1 2016 Q6 [9]}}