| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find velocity from position |
| Difficulty | Moderate -0.3 This is a straightforward vector calculus question requiring differentiation of position to find velocity, then solving a vector equation equal to zero. The techniques are standard (differentiating polynomials, solving simultaneous quadratics) with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{v} = \dfrac{\text{d}\mathbf{r}}{\text{d}t} = (12-2\times2t)\mathbf{i} + (2t-6)\mathbf{j}\) | M1, A1 [2] | Attempt to differentiate at least one coefficient; must use vector notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(t=3\) both components of velocity are zero | M1 | Equating at least one component of their vector velocity to zero; do not allow M1 for solving \(12-4t=2t-6\) unless at least one zero subsequently established |
| so the particle is stationary at \(t=3\) | E1 [2] | Must be argued from two zero components |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{v} = \dfrac{\text{d}\mathbf{r}}{\text{d}t} = (12-2\times2t)\mathbf{i} + (2t-6)\mathbf{j}$ | M1, A1 [2] | Attempt to differentiate at least one coefficient; must use vector notation |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $t=3$ both components of velocity are zero | M1 | Equating at least one component of their vector velocity to zero; do not allow M1 for solving $12-4t=2t-6$ unless at least one zero subsequently established |
| so the particle is stationary at $t=3$ | E1 [2] | Must be argued from two zero components |
---5 The position vector $\mathbf { r }$ metres of a particle at time $t$ seconds is given by
$$\mathbf { r } = \left( 1 + 12 t - 2 t ^ { 2 } \right) \mathbf { i } + \left( t ^ { 2 } - 6 t \right) \mathbf { j }$$
(i) Find an expression for the velocity of the particle at time $t$.\\
(ii) Determine whether the particle is ever stationary.
\hfill \mbox{\textit{OCR MEI Paper 1 2018 Q5 [4]}}