| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Displacement from velocity by integration |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring standard integration techniques twice (acceleration to velocity, velocity to displacement) with given initial conditions, followed by simple substitution. The 'show that' and 'verify' structure makes it more routine than problem-solving, though it does require careful bookkeeping across multiple steps. Slightly easier than average due to its procedural nature and clear guidance. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = \int k(3t^2 - 12t + 2)\ dt = k(3t^3/3 - 12t^2/2 + 2t) + C\) | \*M1 | Use of \(v = \int a\ dt\) |
| \(v = k\left(t^3 - 6t^2 + 2t\right) + C\) | A1 | Condone \(C\) missing |
| \(C = 0.4\) | B1 | |
| \(0.1 = k(1 - 6 + 2) + 0.4\ \ [-0.3 = -3k]\) | DM1 | Substitutes \(t = 1,\ v = 0.1\) |
| \(k = 0.1\) | A1 | AG |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[s = \int 0.1(t^3 - 6t^2 + 2t) + 0.4\ dt = 0.1(t^4/4 - 6t^3/3 + 2t^2/2) + 0.4t + C\right]\) | M1 | Use of \(s = \int v\ dt\) |
| \(s = 0.025t^4 - 0.2t^3 + 0.1t^2 + 0.4t\) | A1 | \(C = 0\) seen or implied |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitutes \(t = 2\) to show \(s = 0\) | B1 | AG |
| 1 |
# Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = \int k(3t^2 - 12t + 2)\ dt = k(3t^3/3 - 12t^2/2 + 2t) + C$ | **\*M1** | Use of $v = \int a\ dt$ |
| $v = k\left(t^3 - 6t^2 + 2t\right) + C$ | **A1** | Condone $C$ missing |
| $C = 0.4$ | **B1** | |
| $0.1 = k(1 - 6 + 2) + 0.4\ \ [-0.3 = -3k]$ | **DM1** | Substitutes $t = 1,\ v = 0.1$ |
| $k = 0.1$ | **A1** | AG |
| | **5** | |
# Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[s = \int 0.1(t^3 - 6t^2 + 2t) + 0.4\ dt = 0.1(t^4/4 - 6t^3/3 + 2t^2/2) + 0.4t + C\right]$ | **M1** | Use of $s = \int v\ dt$ |
| $s = 0.025t^4 - 0.2t^3 + 0.1t^2 + 0.4t$ | **A1** | $C = 0$ seen or implied |
| | **2** | |
# Question 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitutes $t = 2$ to show $s = 0$ | **B1** | AG |
| | **1** | |
---5 A particle starts from a fixed origin with velocity $0.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and moves in a straight line. The acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ of the particle $t \mathrm {~s}$ after it leaves the origin is given by $a = k \left( 3 t ^ { 2 } - 12 t + 2 \right)$, where $k$ is a constant. When $t = 1$, the velocity of $P$ is $0.1 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that the value of $k$ is 0.1 .\\
(ii) Find an expression for the displacement of the particle from the origin in terms of $t$.\\
(iii) Hence verify that the particle is again at the origin at $t = 2$.\\
\hfill \mbox{\textit{CAIE M1 2017 Q5 [8]}}