| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| 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 initial conditions. Students integrate acceleration to get velocity (using the given initial velocity), then integrate velocity to get position (using the given initial position). The algebra is simple, and verifying v=0 at t=3 is routine substitution. This is slightly easier than average because it's a standard textbook exercise with no conceptual challenges beyond basic calculus application. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = \int(6t - 12)\,\mathrm{d}t\) | M1 | Attempt to integrate |
| \(v = 3t^2 - 12t + c\) | A1 | Condone no \(c\) if implied by subsequent working |
| \(c = 9\) | A1 | |
| \(t = 3 \Rightarrow v = 3 \times 3^2 - 12 \times 3 + 9 = 0\) | E1 | Or by showing \((t-3)\) is a factor of \(3t^2 - 12t + 9\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s = \int(3t^2 - 12t + 9)\,\mathrm{d}t\) | 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 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = \int(6t - 12)\,\mathrm{d}t$ | M1 | Attempt to integrate |
| $v = 3t^2 - 12t + c$ | A1 | Condone no $c$ if implied by subsequent working |
| $c = 9$ | A1 | |
| $t = 3 \Rightarrow v = 3 \times 3^2 - 12 \times 3 + 9 = 0$ | E1 | Or by showing $(t-3)$ is a factor of $3t^2 - 12t + 9$ |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s = \int(3t^2 - 12t + 9)\,\mathrm{d}t$ | 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 |3 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 Q3 [7]}}