| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Variable acceleration with initial conditions |
| Difficulty | Standard +0.3 This is a standard variable acceleration question requiring integration of a linear acceleration function with two initial conditions to find constants. The steps are routine: integrate twice, apply boundary conditions, solve for constants. Part (i) requires understanding that deceleration means velocity and acceleration have opposite signs, which adds slight conceptual depth beyond pure calculation, but overall this is a straightforward textbook exercise 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 |
| \([6t - 2 < 0 \rightarrow t < .....]\) | M1 | For solving \(a(t) < 0\) |
| \(0 < t < \frac{1}{3}\) | A1 | [2 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([v = 3t^2 - 2t + c]\) | M1 | For using \(v(t) = \int a(t)\,dt\) |
| M1 | For using \(s(t) = \int v(t)\,dt\) | |
| \(s = t^3 - t^2 + ct + d\) | A1 | |
| \([c + d = 7\); \(3c + d = 11 \rightarrow c = ..., d = ...]\) | M1 | For using \(t=1\), \(s=7\) and \(t=3\), \(s=29\) to form and solve simultaneous equations |
| \(s = t^3 - t^2 + 2t + 5\) | A1 | [5 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([3t^2 - 2t + 2 = 10]\) | M1 | For using \(v(t) = 10\) |
| DM1 | For solving 3 term quadratic \(v(t) = 10\) | |
| \(t = 2\) | A1 | [3 marks] |
## Question 7:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[6t - 2 < 0 \rightarrow t < .....]$ | M1 | For solving $a(t) < 0$ |
| $0 < t < \frac{1}{3}$ | A1 | **[2 marks]** |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[v = 3t^2 - 2t + c]$ | M1 | For using $v(t) = \int a(t)\,dt$ |
| | M1 | For using $s(t) = \int v(t)\,dt$ |
| $s = t^3 - t^2 + ct + d$ | A1 | |
| $[c + d = 7$; $3c + d = 11 \rightarrow c = ..., d = ...]$ | M1 | For using $t=1$, $s=7$ and $t=3$, $s=29$ to form and solve simultaneous equations |
| $s = t^3 - t^2 + 2t + 5$ | A1 | **[5 marks]** |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[3t^2 - 2t + 2 = 10]$ | M1 | For using $v(t) = 10$ |
| | DM1 | For solving 3 term quadratic $v(t) = 10$ |
| $t = 2$ | A1 | **[3 marks]** |7 A particle $P$ moves in a straight line. At time $t \mathrm {~s}$, the displacement of $P$ from $O$ is $s \mathrm {~m}$ and the acceleration of $P$ is $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, where $a = 6 t - 2$. When $t = 1 , s = 7$ and when $t = 3 , s = 29$.\\
(i) Find the set of values of $t$ for which the particle is decelerating.\\
(ii) Find $s$ in terms of $t$.\\
(iii) Find the time when the velocity of the particle is $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\hfill \mbox{\textit{CAIE M1 2016 Q7 [10]}}