| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Moderate -0.8 This is a straightforward M1 kinematics question requiring standard differentiation to find acceleration, solving a quadratic for zero velocity, and integration for distance. All techniques are routine with no problem-solving insight needed, making it easier than average but not trivial due to the multi-step integration component. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
2 A particle moves along the $x$-axis with velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, at time $t$ given by
$$v = 24 t - 6 t ^ { 2 }$$
The positive direction is in the sense of $x$ increasing.\\
(i) Find an expression for the acceleration of the particle at time $t$.\\
(ii) Find the times, $t _ { 1 }$ and $t _ { 2 }$, at which the particle has zero speed.\\
(iii) Find the distance travelled between the times $t _ { 1 }$ and $t _ { 2 }$.
\hfill \mbox{\textit{OCR MEI M1 2005 Q2 [8]}}