| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Moderate -0.3 This is a standard M1 kinematics question involving integration of a quadratic velocity function. Parts (i)-(iii) are routine substitution and differentiation. Parts (iv)-(vi) require integrating to find displacement and accounting for direction changes, which is a core M1 skill but straightforward once the stationary points are identified. The question is slightly easier than average due to its structured, multi-part nature that guides students through the solution. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = 0 - 0 - 8 = -8\) m s\(^{-1}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(t^2 - 2t - 8 = 0 \Rightarrow (t-4)(t+2) = 0\) | M1 | Factorise or solve |
| \(t = 4\) or \(t = -2\) | A1 | Both values confirmed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dv}{dt} = 2t - 2 = 0 \Rightarrow t = 1\) | M1 A1 | Differentiate and set to zero |
| \(v = 1 - 2 - 8 = -9\) m s\(^{-1}\) | A1 | |
| Point A has coordinates \((1, -9)\) | B1 B1 | Both coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_1^4 (t^2-2t-8)\,dt = \left[\frac{t^3}{3}-t^2-8t\right]_1^4\) | M1 | Integrate |
| \(= \left(\frac{64}{3}-16-32\right) - \left(\frac{1}{3}-1-8\right)\) | A1 | Correct integration |
| \(= -\frac{100}{3} - (-\frac{24}{3}) = -\frac{76}{3}\) | A1 | |
| Distance \(= \frac{76}{3} \approx 25.3\) m | A1 | Magnitude (negative means opposite direction) |
| (Note: check if sign change needed within interval) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distance \(= \frac{76}{3}\) m (from symmetry/calculation, as \(v\leq0\) throughout \(-2\leq t\leq4\)) | B1 | ft from (iv) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_4^5 (t^2-2t-8)\,dt = \left[\frac{t^3}{3}-t^2-8t\right]_4^5\) | M1 | |
| \(= \left(\frac{125}{3}-25-40\right)-\left(\frac{64}{3}-16-32\right) = \frac{61}{3}-(-\frac{80}{3})\) | A1 | |
| Total distance from \(t=1\): distance from 1 to 4 plus distance from 4 to 5 | M1 | |
| Distance from 4 to 5 \(= \frac{7}{3}\) m; total \(= \frac{76}{3}+\frac{7}{3} = \frac{83}{3} \approx 27.7\) m | A1 | Wait — question asks only \(1\leq t\leq5\); answer \(\approx 27.7\) m |
# Question 7:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = 0 - 0 - 8 = -8$ m s$^{-1}$ | B1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $t^2 - 2t - 8 = 0 \Rightarrow (t-4)(t+2) = 0$ | M1 | Factorise or solve |
| $t = 4$ or $t = -2$ | A1 | Both values confirmed |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dv}{dt} = 2t - 2 = 0 \Rightarrow t = 1$ | M1 A1 | Differentiate and set to zero |
| $v = 1 - 2 - 8 = -9$ m s$^{-1}$ | A1 | |
| Point A has coordinates $(1, -9)$ | B1 B1 | Both coordinates |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_1^4 (t^2-2t-8)\,dt = \left[\frac{t^3}{3}-t^2-8t\right]_1^4$ | M1 | Integrate |
| $= \left(\frac{64}{3}-16-32\right) - \left(\frac{1}{3}-1-8\right)$ | A1 | Correct integration |
| $= -\frac{100}{3} - (-\frac{24}{3}) = -\frac{76}{3}$ | A1 | |
| Distance $= \frac{76}{3} \approx 25.3$ m | A1 | Magnitude (negative means opposite direction) |
| (Note: check if sign change needed within interval) | | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distance $= \frac{76}{3}$ m (from symmetry/calculation, as $v\leq0$ throughout $-2\leq t\leq4$) | B1 | ft from (iv) |
## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_4^5 (t^2-2t-8)\,dt = \left[\frac{t^3}{3}-t^2-8t\right]_4^5$ | M1 | |
| $= \left(\frac{125}{3}-25-40\right)-\left(\frac{64}{3}-16-32\right) = \frac{61}{3}-(-\frac{80}{3})$ | A1 | |
| Total distance from $t=1$: distance from 1 to 4 plus distance from 4 to 5 | M1 | |
| Distance from 4 to 5 $= \frac{7}{3}$ m; total $= \frac{76}{3}+\frac{7}{3} = \frac{83}{3} \approx 27.7$ m | A1 | Wait — question asks only $1\leq t\leq5$; answer $\approx 27.7$ m |
---7 Fig. 7 is a sketch of part of the velocity-time graph for the motion of an insect walking in a straight line. Its velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, at time $t$ seconds for the time interval $- 3 \leqslant t \leqslant 5$ is given by
$$v = t ^ { 2 } - 2 t - 8 .$$
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3be85526-3872-42ac-8278-1d4a3cf75ff7-5_646_898_552_587}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
(i) Write down the velocity of the insect when $t = 0$.\\
(ii) Show that the insect is instantaneously at rest when $t = - 2$ and when $t = 4$.\\
(iii) Determine the velocity of the insect when its acceleration is zero.
Write down the coordinates of the point A shown in Fig. 7.\\
(iv) Calculate the distance travelled by the insect from $t = 1$ to $t = 4$.\\
(v) Write down the distance travelled by the insect in the time interval $- 2 \leqslant t \leqslant 4$.\\
(vi) How far does the insect walk in the time interval $1 \leqslant t \leqslant 5$ ?
\hfill \mbox{\textit{OCR MEI M1 2007 Q7 [17]}}