| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration from velocity differentiation |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring basic differentiation to find acceleration and integration to find displacement. Both operations involve simple polynomial functions with no conceptual challenges—students apply standard formulas directly without problem-solving or geometric insight. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Acceleration is \(1 + 0.2t\) | M1, A1 [2] | For using \(a = \dot{v}(t)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(t = 9\) | M1, A1 | For solving \(a(t) = 2.8\) for \(t\) |
| M1* | For integrating \(v(t)\) to find \(s(t)\) | |
| A1 | For \(t^2 \div 2\) correct in \(s(t)\) | |
| A1 | For \(t^3 \div 30\) correct in \(s(t)\) | |
| \(s(9) = 9^2 \div 2 + 9^3 \div 30 - (0 + 0)\) \((= 40.5 + 24.3)\) | dep*M1 | For correct use of limits or equivalent |
| Distance is \(64.8\) m | A1ft [7] | ft their \(a = \dot{v}(t)\) from (i) |
# Question 4:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Acceleration is $1 + 0.2t$ | M1, A1 [2] | For using $a = \dot{v}(t)$ |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $t = 9$ | M1, A1 | For solving $a(t) = 2.8$ for $t$ |
| | M1* | For integrating $v(t)$ to find $s(t)$ |
| | A1 | For $t^2 \div 2$ correct in $s(t)$ |
| | A1 | For $t^3 \div 30$ correct in $s(t)$ |
| $s(9) = 9^2 \div 2 + 9^3 \div 30 - (0 + 0)$ $(= 40.5 + 24.3)$ | dep*M1 | For correct use of limits or equivalent |
| Distance is $64.8$ m | A1ft [7] | ft their $a = \dot{v}(t)$ from (i) |
---4 A particle moves in a straight line. Its velocity $t \mathrm {~s}$ after leaving a fixed point on the line is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where $v = t + 0.1 t ^ { 2 }$. Find\\
(i) an expression for the acceleration of the particle at time $t$,\\
(ii) the distance travelled by the particle from time $t = 0$ until the instant when its acceleration is $2.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\hfill \mbox{\textit{OCR M1 2005 Q4 [9]}}