| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Variable acceleration with initial conditions |
| Difficulty | Moderate -0.3 This is a straightforward integration problem with given initial conditions. Students integrate acceleration to get velocity (using v = 9 at t = 0), verify a specific value by substitution, then integrate again to find position (using s = -2 at t = 0). It requires only standard calculus techniques with no problem-solving insight, making it slightly easier than average but not trivial due to the two-stage integration process. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = \int (6t-12)\,dt\) | M1 | Attempt to integrate |
| \(v = 3t^2 - 12t + c\) | A1 | Condone no \(c\) if implied by subsequent working |
| \(c = 9\) | A1 | |
| \(t=3 \Rightarrow v = 3\times3^2 - 12\times3 + 9 = 0\) | E1 | Or by showing that \((t-3)\) is a factor of \(3t^2-12t+9\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s = \int(3t^2-12t+9)\,dt\) | M1 | Attempt to integrate. Ft from part (i) |
| \(s = t^3 - 6t^2 + 9t - 2\) | A1 | A correct value of \(c\) is required. Ft from part (i) |
| When \(t=2\), \(s=0\) (It is at the origin.) | B1 | Cao |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = \int (6t-12)\,dt$ | M1 | Attempt to integrate |
| $v = 3t^2 - 12t + c$ | A1 | Condone no $c$ if implied by subsequent working |
| $c = 9$ | A1 | |
| $t=3 \Rightarrow v = 3\times3^2 - 12\times3 + 9 = 0$ | E1 | Or by showing that $(t-3)$ is a factor of $3t^2-12t+9$ |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s = \int(3t^2-12t+9)\,dt$ | M1 | Attempt to integrate. Ft from part (i) |
| $s = t^3 - 6t^2 + 9t - 2$ | A1 | A correct value of $c$ is required. Ft from part (i) |
| When $t=2$, $s=0$ (It is at the origin.) | B1 | Cao |
---2 A particle is moving along a straight line and its position is relative to an origin on the line. At time $t \mathrm {~s}$, the particle's acceleration, $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, is given by
$$a = 6 t - 12$$
At $t = 0$ the velocity of the particle is $+ 9 \mathrm {~ms} ^ { - 1 }$ and its position is - 2 m .\\
(i) Find an expression for the velocity of the particle at time $t \mathrm {~s}$ and verify that it is stationary when $t = 3$.\\
(ii) Find the position of the particle when $t = 2$.
\hfill \mbox{\textit{OCR MEI M1 2012 Q2 [7]}}