| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Displacement from velocity by integration |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring basic calculus skills: setting v=0 and solving a quadratic for part (i), then integrating v to find x(t) and substituting values in part (ii). All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = 0 \Rightarrow 3(t-2)(t-4) = 0\) | M1 | Setting \(v = 0\) (may be implied) |
| \(T_1 = 2,\ T_2 = 4\) | A1 | Accept \(t = 2\) and \(t = 4\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = \int v\, dt\) | M1 | Use of integration |
| \(x = 24t - 9t^2 + t^3 + c;\ c = 0\) | A1 | Condone omission of \(c\) |
| \(t = 2 \Rightarrow x = 48 - 36 + 8 = 20\) | E1 | CAO |
| \(t = 4 \Rightarrow x = 96 - 144 + 64 = 16\) | A1 | CAO |
| [4] |
## Question 4:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = 0 \Rightarrow 3(t-2)(t-4) = 0$ | M1 | Setting $v = 0$ (may be implied) |
| $T_1 = 2,\ T_2 = 4$ | A1 | Accept $t = 2$ and $t = 4$ |
| **[2]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \int v\, dt$ | M1 | Use of integration |
| $x = 24t - 9t^2 + t^3 + c;\ c = 0$ | A1 | Condone omission of $c$ |
| $t = 2 \Rightarrow x = 48 - 36 + 8 = 20$ | E1 | CAO |
| $t = 4 \Rightarrow x = 96 - 144 + 64 = 16$ | A1 | CAO |
| **[4]** | | |4 A particle moves along a straight line through an origin O . Initially the particle is at O .\\
At time $t \mathrm {~s}$, its displacement from O is $x \mathrm {~m}$ and its velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, is given by
$$v = 24 - 18 t + 3 t ^ { 2 }$$
(i) Find the times, $T _ { 1 } \mathrm {~s}$ and $T _ { 2 } \mathrm {~s}$ (where $T _ { 2 } > T _ { 1 }$ ), at which the particle is stationary.\\
(ii) Find an expression for $x$ at time $t$ s.
Show that when $t = T _ { 1 } , x = 20$ and find the value of $x$ when $t = T _ { 2 }$.
\hfill \mbox{\textit{OCR MEI M1 Q4 [6]}}