| Exam Board | OCR MEI |
|---|---|
| Module | Further Numerical Methods (Further Numerical Methods) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule applied to real-world data |
| Difficulty | Standard +0.3 This is a straightforward application of Newton's interpolating polynomial with standard numerical integration. Part (i) requires basic physics recall (constant force → constant acceleration → quadratic velocity), part (ii) is a routine algorithm from the syllabus, parts (iii-iv) are standard interpretation/integration, and part (v) asks for method identification. All steps are procedural with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.09f Trapezium rule: numerical integration |
| Time \(( t\) seconds \()\) | 5 | 10 | 12 | 15 |
| Velocity \(( v\) metres per second \()\) | 5.1250 | 11.000 | 14.000 | 18.375 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (i) | [net] force is a linear function [of time] |
| [1] | 2.4 | |
| 5 | (ii) | Uses values at 5, 10, 15 |
| Answer | Marks |
|---|---|
| v(cid:32)0.03t2 (cid:14)0.725t(cid:14)0.75 | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.3 |
| Answer | Marks |
|---|---|
| 3.3 | soi N |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (iii) | f(12) = 13.77 |
| Answer | Marks |
|---|---|
| reasonable approximation | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3.5a | M | |
| 5 | (iv) | E |
| Answer | Marks |
|---|---|
| 112.5 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| 1.1 | Attempt at integration must be |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (v) | P |
| Answer | Marks | Guidance |
|---|---|---|
| Lagrange's form of the interpolating polynomial | B1 | |
| [1] | 3.5c | accept |
Question 5:
5 | (i) | [net] force is a linear function [of time] | B1
[1] | 2.4
5 | (ii) | Uses values at 5, 10, 15
First differences: 5.875, 7.375
Second difference: 1.5
(t(cid:16)5) (t(cid:16)5)(t(cid:16)10)
5.125(cid:14) (cid:117)5.875(cid:14) (cid:117)1.5
5 52(cid:117)2!
v(cid:32)0.03t2 (cid:14)0.725t(cid:14)0.75 | M1
M1
M1
A1
[4] | 3.3
1.2
1.1
3.3 | soi N
E
FT differences
5 | (iii) | f(12) = 13.77
Quite close to 14.000, so the model may be a
reasonable approximation | B1
C
E1
[2] | 1.1
I
3.5a | M
5 | (iv) | E
15
15 (cid:170)0.03t3 0.725t2 (cid:186)
(cid:179) vdt(cid:32)(cid:171) (cid:14) (cid:14)0.75t(cid:187)
5 (cid:172) 3 2 (cid:188)
5
112.5 | M1
A1
[2] | 3.4
1.1 | Attempt at integration must be
seen.
5 | (v) | P
S
Lagrange's form of the interpolating polynomial | B1
[1] | 3.5c | accept
(x(cid:16)5)(x(cid:16)10)(x(cid:16)15)
(cid:117)14.000
(12(cid:16)5)(12(cid:16)10)(12(cid:16)15)5 A vehicle is moving in a straight line. Its velocity at different times is recorded and shown below. The velocities are recorded to 5 significant figures and the times may be assumed to be exact.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Time $( t$ seconds $)$ & 5 & 10 & 12 & 15 \\
\hline
Velocity $( v$ metres per second $)$ & 5.1250 & 11.000 & 14.000 & 18.375 \\
\hline
\end{tabular}
\end{center}
It is suggested initially that a quadratic model may be appropriate for this situation.\\
(i) Given that the vehicle is modelled as a particle with constant mass, what assumption about the net force acting on the vehicle leads to a quadratic model?\\
(ii) Find Newton's interpolating polynomial of degree 2 to model this situation. Write your answer in the form $v = a t ^ { 2 } + b t + c$.\\
(iii) Comment on whether this model appears to be appropriate.\\
(iv) Use this model to find an approximation to the distance travelled over the interval $5 \leq t \leq 15$.
Further investigation suggests that a cubic model may be more appropriate.\\
(v) What technique would you use to fit a cubic model to the data in the table?
\hfill \mbox{\textit{OCR MEI Further Numerical Methods Q5 [10]}}