Question 3 - A-Level Maths

OCR MEI M1 2009 June — Question 3 8 marks

Exam Board	OCR MEI
Module	M1 (Mechanics 1)
Year	2009
Session	June
Marks	8
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 differentiation of a quadratic to find acceleration, interpretation of zero acceleration, and integration of a polynomial for distance. All techniques are routine M1 content with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure.
Spec	1.08d Evaluate definite integrals: between limits 3.02f Non-uniform acceleration: using differentiation and integration

3 Fig. 3 is a sketch of the velocity-time graph modelling the velocity of a sprinter at the start of a race. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{d6e78f93-ac2c-4053-87e4-5e5537d6dc3d-3_588_1091_351_529} \captionsetup{labelformat=empty} \caption{Fig. 3}

\end{figure}

How can you tell from the sketch that the acceleration is not modelled as being constant for $0 \leqslant t \leqslant 4$ ? The velocity of the sprinter, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, for the time interval $0 \leqslant t \leqslant 4$ is modelled by the expression $$v = 3 t - \frac { 3 } { 8 } t ^ { 2 }$$
Find the acceleration that the model predicts for $t = 4$ and comment on what this suggests about the running of the sprinter.
Calculate the distance run by the sprinter from $t = 1$ to $t = 4$.

Show mark scheme Show mark scheme source

Question 3:

Part (i)

Answer	Marks	Guidance
The line is not straight	B1	Any valid comment

Part (ii)

Answer	Marks	Guidance
$a = 3 - \frac{6t}{8}$	M1	Attempt to differentiate. Accept 1 term correct but not $3 - \frac{3t}{8}$
$a(4) = 0$	F1
The sprinter has reached a steady speed	E1	Accept 'stopped accelerating' but not just $a = 0$. Do not FT $a(4) \neq 0$

Part (iii)

Answer	Marks	Guidance
We require $\int_{1}^{4}\left(3t - \frac{3t^2}{8}\right)dt$	M1	Integrating. Neglect limits
$= \left[\frac{3t^2}{2} - \frac{t^3}{8}\right]_{1}^{4}$	A1	One term correct. Neglect limits
$= (24-8) - \left(\frac{3}{2} - \frac{1}{8}\right)$	M1	Correct limits substituted in integral. Subtraction seen. If arb constant used, evaluated to give $s=0$ when $t=1$ and then sub $t=4$
$= 14\frac{5}{8}$ m (14.625 m)	A1	cao. Any form. [If trapezium rule used: M1 use of rule (must be clear method and at least two regions), A1 correctly applied, M1 at least 6 regions used, A1 answer correct to at least 2 s.f.]

# Question 3:

## Part (i)
| The line is not straight | B1 | Any valid comment |

## Part (ii)
| $a = 3 - \frac{6t}{8}$ | M1 | Attempt to differentiate. Accept 1 term correct but not $3 - \frac{3t}{8}$ |
| $a(4) = 0$ | F1 | |
| The sprinter has reached a steady speed | E1 | Accept 'stopped accelerating' but not just $a = 0$. Do not FT $a(4) \neq 0$ |

## Part (iii)
| We require $\int_{1}^{4}\left(3t - \frac{3t^2}{8}\right)dt$ | M1 | Integrating. Neglect limits |
| $= \left[\frac{3t^2}{2} - \frac{t^3}{8}\right]_{1}^{4}$ | A1 | One term correct. Neglect limits |
| $= (24-8) - \left(\frac{3}{2} - \frac{1}{8}\right)$ | M1 | Correct limits substituted in integral. Subtraction seen. If arb constant used, evaluated to give $s=0$ when $t=1$ and then sub $t=4$ |
| $= 14\frac{5}{8}$ m (14.625 m) | A1 | cao. Any form. [If trapezium rule used: M1 use of rule (must be clear method and at least two regions), A1 correctly applied, M1 at least 6 regions used, A1 answer correct to at least 2 s.f.] |

---

Show LaTeX source

3 Fig. 3 is a sketch of the velocity-time graph modelling the velocity of a sprinter at the start of a race.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{d6e78f93-ac2c-4053-87e4-5e5537d6dc3d-3_588_1091_351_529}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}

(i) How can you tell from the sketch that the acceleration is not modelled as being constant for $0 \leqslant t \leqslant 4$ ?

The velocity of the sprinter, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, for the time interval $0 \leqslant t \leqslant 4$ is modelled by the expression

$$v = 3 t - \frac { 3 } { 8 } t ^ { 2 }$$

(ii) Find the acceleration that the model predicts for $t = 4$ and comment on what this suggests about the running of the sprinter.\\
(iii) Calculate the distance run by the sprinter from $t = 1$ to $t = 4$.

\hfill \mbox{\textit{OCR MEI M1 2009 Q3 [8]}}

This paper (6 questions)

View full paper

Q1 6 Q2 7 Q3 8 Q4 7 Q5 8 Q6 16

Answer	Marks	Guidance
\(a = 3 - \frac{6t}{8}\)	M1	Attempt to differentiate. Accept 1 term correct but not \(3 - \frac{3t}{8}\)
\(a(4) = 0\)	F1
The sprinter has reached a steady speed	E1	Accept 'stopped accelerating' but not just \(a = 0\). Do not FT \(a(4) \neq 0\)

Answer	Marks	Guidance
We require \(\int_{1}^{4}\left(3t - \frac{3t^2}{8}\right)dt\)	M1	Integrating. Neglect limits
\(= \left[\frac{3t^2}{2} - \frac{t^3}{8}\right]_{1}^{4}\)	A1	One term correct. Neglect limits
\(= (24-8) - \left(\frac{3}{2} - \frac{1}{8}\right)\)	M1	Correct limits substituted in integral. Subtraction seen. If arb constant used, evaluated to give \(s=0\) when \(t=1\) and then sub \(t=4\)
\(= 14\frac{5}{8}\) m (14.625 m)	A1	cao. Any form. [If trapezium rule used: M1 use of rule (must be clear method and at least two regions), A1 correctly applied, M1 at least 6 regions used, A1 answer correct to at least 2 s.f.]