| 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 M1 kinematics question requiring basic calculus operations: differentiate to find acceleration, solve a quadratic for zero velocity, and integrate to find distance. All techniques are standard with no conceptual challenges or novel problem-solving required, making it easier than average but not trivial due to the integration step. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| 2(i) | a=24−12t | M1 |
| A1 | Differentiate | |
| cao | 2 | |
| (ii) | Need 24t−6t2 =0 | |
| t = 0, 4 | M1 | |
| A1 | Equate v = 0 and attempt to factorise |
| Answer | Marks |
|---|---|
| Both. cao. | 2 |
| (iii) | 4 |
| Answer | Marks |
|---|---|
| = 64 m | M1 |
| Answer | Marks |
|---|---|
| A1 | Attempt to integrate. No limits required. |
| Answer | Marks |
|---|---|
| A1 (dep on 2nd M1) One strip area correct: A1 cao] | 4 |
| total | 8 |
Question 2:
--- 2(i) ---
2(i) | a=24−12t | M1
A1 | Differentiate
cao | 2
(ii) | Need 24t−6t2 =0
t = 0, 4 | M1
A1 | Equate v = 0 and attempt to factorise
(or solve). Award for one root found.
Both. cao. | 2
(iii) | 4
( 24t−6t2)
s=∫ dt
0
= ⎡12t2 −2t3⎤4
⎣ ⎦
0
(12×16−2×64)−0
= 64 m | M1
A1
M1
A1 | Attempt to integrate. No limits required.
Either term correct. No limits required
Sub t = 4 in integral. Accept no bottom limit
substituted or arb const assumed 0. Accept reversed
limits. FT their limits.
cao. Award if seen.
[If trapezium rule used.
M1 At least 4 strips: M1 enough strips for 3 s. f.
A1 (dep on 2nd M1) One strip area correct: A1 cao] | 4
total | 8A particle moves along the $x$-axis with velocity, $v$ ms$^{-1}$, at time $t$ given by
$$v = 24t - 6t^2.$$
The positive direction is in the sense of $x$ increasing.
\begin{enumerate}[label=(\roman*)]
\item Find an expression for the acceleration of the particle at time $t$. [2]
\item Find the times, $t_1$ and $t_2$, at which the particle has zero speed. [2]
\item Find the distance travelled between the times $t_1$ and $t_2$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 Q2 [8]}}