| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration from velocity differentiation |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring standard differentiation to find acceleration from velocity (part i) and integration to find displacement from velocity (part ii). Both are routine A-level mechanics techniques with no conceptual challenges—the chain rule for differentiation and power rule for integration are applied directly to the given cubic expression. Slightly easier than average due to the mechanical nature of the calculations. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = 3 \times 2 \times (2t-5)^2\ [=54]\) | *M1 | Uses \(a = dv/dt\) |
| \(6(2t-5)^2 = 54 \rightarrow t = \ldots\) | DM1 | Solves for \(t\) |
| \(t = 1, 4\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s = \frac{(2t-5)^4}{4 \times 2}\) \((+C)\) | \*M1 | Uses \(s = \int v \, dt\) |
| \(C = -\frac{625}{8}\) | DM1 | Uses \(s = 0\) at \(t = 0\) |
| \(s = \frac{(2t-5)^4}{8} - \frac{625}{8}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = 8t^3 - 60t^2 + 150t - 125 \rightarrow a = 24t^2 - 120t + 150\) | \*M1 | Uses \(a = dv/dt\) |
| \(24t^2 - 120t + 150 = 54 \rightarrow t = \ldots\) | DM1 | Solves for \(t\) |
| \(t = 1, 4\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s = \int 8t^3 - 60t^2 + 150t - 125 \, dt \rightarrow s = \frac{8}{4}t^4 - \frac{60}{3}t^3 + \frac{150}{2}t^2 - 125t\) \((+C)\) | \*M1 | Uses \(s = \int v \, dt\) |
| \(C = 0\) | DM1 | Uses \(s = 0\) at \(t = 0\) (may be implied) |
| \(s = 2t^4 - 20t^3 + 75t^2 - 125t\) | A1 |
## Question 4:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = 3 \times 2 \times (2t-5)^2\ [=54]$ | *M1 | Uses $a = dv/dt$ |
| $6(2t-5)^2 = 54 \rightarrow t = \ldots$ | DM1 | Solves for $t$ |
| $t = 1, 4$ | A1 | |
| **Total: 3** | | |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s = \frac{(2t-5)^4}{4 \times 2}$ $(+C)$ | \*M1 | Uses $s = \int v \, dt$ |
| $C = -\frac{625}{8}$ | DM1 | Uses $s = 0$ at $t = 0$ |
| $s = \frac{(2t-5)^4}{8} - \frac{625}{8}$ | A1 | |
**Alternative method for Question 4:**
## Question 4(i) [Alternative]:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = 8t^3 - 60t^2 + 150t - 125 \rightarrow a = 24t^2 - 120t + 150$ | \*M1 | Uses $a = dv/dt$ |
| $24t^2 - 120t + 150 = 54 \rightarrow t = \ldots$ | DM1 | Solves for $t$ |
| $t = 1, 4$ | A1 | |
## Question 4(ii) [Alternative]:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s = \int 8t^3 - 60t^2 + 150t - 125 \, dt \rightarrow s = \frac{8}{4}t^4 - \frac{60}{3}t^3 + \frac{150}{2}t^2 - 125t$ $(+C)$ | \*M1 | Uses $s = \int v \, dt$ |
| $C = 0$ | DM1 | Uses $s = 0$ at $t = 0$ (may be implied) |
| $s = 2t^4 - 20t^3 + 75t^2 - 125t$ | A1 | |
---4 A particle $P$ moves in a straight line starting from a point $O$. At time $t \mathrm {~s}$ after leaving $O$, the velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, of $P$ is given by $v = ( 2 t - 5 ) ^ { 3 }$.\\
(i) Find the values of $t$ when the acceleration of $P$ is $54 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(ii) Find an expression for the displacement of $P$ from $O$ at time $t \mathrm {~s}$.\\
\hfill \mbox{\textit{CAIE M1 2017 Q4 [6]}}