| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity function using calculus |
| Difficulty | Moderate -0.8 This is a straightforward mechanics calculus question requiring only standard differentiation of a polynomial for acceleration and integration for displacement. All three parts are routine applications of basic techniques with no problem-solving insight needed, making it easier than average for A-level. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
The velocity, $v$ ms$^{-1}$, of a particle moving along a straight line is given by
$$v = 3t^2 - 12t + 14,$$
where $t$ is the time in seconds.
\begin{enumerate}[label=(\roman*)]
\item Find an expression for the acceleration of the particle at time $t$. [2]
\item Find the displacement of the particle from its position when $t = 1$ to its position when $t = 3$. [4]
\item You are given that $v$ is always positive. Explain how this tells you that the distance travelled by the particle between $t = 1$ and $t = 3$ has the same value as the displacement between these times. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 Q5 [8]}}