| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| 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 and interpretation of velocity-time graphs. While it requires multiple steps (finding roots, integrating with attention to sign changes), all techniques are routine: substitution into quadratic, factorizing, basic integration of polynomials, and calculating areas with direction changes. The multi-part structure guides students through the problem systematically, making it slightly easier than average despite the length. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.08e Area between curve and x-axis: using definite integrals3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(8\ \text{m s}^{-1}\) (in the negative direction) | B1 | Allow \(\pm\) and no direction indicated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((t+2)(t-4) = 0\) so \(t = -2\) or \(4\) | M1 | Equating \(v\) to zero and solving or substituting |
| A1 | If subst used then both must be clearly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 2t - 2\) | M1 | Differentiating |
| \(a = 2t - 2\) (correct) | A1 | |
| \(a = 0\) when \(t = 1\) | F1 | |
| \(v(1) = 1 - 2 - 8 = -9\) | ||
| so \(9\ \text{m s}^{-1}\) in the negative direction | A1 | Accept \(-9\) but not \(9\) without comment |
| \((1,\ -9)\) | B1 | FT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_1^4 (t^2 - 2t - 8)\,dx\) | M1 | Attempt at integration. Ignore limits |
| \(= \left[\frac{t^3}{3} - t^2 - 8t\right]_1^4\) | A1 | Correct integration. Ignore limits |
| \(= \left(\frac{64}{3} - 16 - 32\right) - \left(\frac{1}{3} - 1 - 8\right)\) | M1 | Attempt to sub correct limits and subtract |
| \(= -18\) | A1 | Limits correctly evaluated. Award if \(-18\) seen but no need to evaluate. Award even if \(-18\) not seen. Do not award for \(-18\) |
| distance is \(18\) m | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2 \times 18 = 36\) m | F1 | Award for \(2 \times\) their (iv) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_4^5 (t^2 - 2t - 8)\,dx = \left[\frac{t^3}{3} - t^2 - 8t\right]_4^5\) | M1 | \(\int_4^5\) attempted or otherwise complete method seen |
| \(= \left(\frac{125}{3} - 25 - 40\right) - \left(-\frac{80}{3}\right) = 3\frac{1}{3}\) | A1 | Correct substitution |
| so \(3\frac{1}{3} + 18 = 21\frac{1}{3}\) m | A1 | Award for \(3\frac{1}{3}\) + their (positive) (iv) |
## Question 6:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $8\ \text{m s}^{-1}$ (in the negative direction) | B1 | Allow $\pm$ and no direction indicated |
**Subtotal: 1 mark**
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(t+2)(t-4) = 0$ so $t = -2$ or $4$ | M1 | Equating $v$ to zero and solving or substituting |
| | A1 | If subst used then both must be clearly shown |
**Subtotal: 2 marks**
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 2t - 2$ | M1 | Differentiating |
| $a = 2t - 2$ (correct) | A1 | |
| $a = 0$ when $t = 1$ | F1 | |
| $v(1) = 1 - 2 - 8 = -9$ | | |
| so $9\ \text{m s}^{-1}$ in the negative direction | A1 | Accept $-9$ but not $9$ without comment |
| $(1,\ -9)$ | B1 | FT |
**Subtotal: 5 marks**
### Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_1^4 (t^2 - 2t - 8)\,dx$ | M1 | Attempt at integration. Ignore limits |
| $= \left[\frac{t^3}{3} - t^2 - 8t\right]_1^4$ | A1 | Correct integration. Ignore limits |
| $= \left(\frac{64}{3} - 16 - 32\right) - \left(\frac{1}{3} - 1 - 8\right)$ | M1 | Attempt to sub correct limits and subtract |
| $= -18$ | A1 | Limits correctly evaluated. Award if $-18$ seen but no need to evaluate. Award even if $-18$ not seen. Do not award for $-18$ |
| distance is $18$ m | A1 | cao |
**Subtotal: 5 marks**
### Part (v)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2 \times 18 = 36$ m | F1 | Award for $2 \times$ **their** (iv) |
**Subtotal: 1 mark**
### Part (vi)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_4^5 (t^2 - 2t - 8)\,dx = \left[\frac{t^3}{3} - t^2 - 8t\right]_4^5$ | M1 | $\int_4^5$ attempted or otherwise complete method seen |
| $= \left(\frac{125}{3} - 25 - 40\right) - \left(-\frac{80}{3}\right) = 3\frac{1}{3}$ | A1 | Correct substitution |
| so $3\frac{1}{3} + 18 = 21\frac{1}{3}$ m | A1 | Award for $3\frac{1}{3}$ + **their** (positive) (iv) |
**Subtotal: 3 marks**
**Total: 17 marks**6 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]{34e4ce80-21b0-48f5-865c-de4dd837f7c5-5_624_886_549_631}
\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 Q6 [17]}}