| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Read and interpret velocity-time graph |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics question testing routine application of kinematics: calculating areas under velocity-time graphs, using SUVAT equations, differentiating to find acceleration, and integrating velocity to find displacement. All parts follow textbook procedures with no novel problem-solving required, though the multi-part structure and integration of a cubic function elevate it slightly above the most basic exercises. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02c Interpret kinematic graphs: gradient and area3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.5 \times 2 \times 12 + 0.5 \times 4 \times 12\) so \(36\) m | M1, A1 | Attempt at sum of areas or equivalent. No extra areas. |
| Answer | Marks | Guidance |
|---|---|---|
| \(8 - \frac{36}{12} = 5\) seconds | B1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(-6\) m s\(^{-2}\) | M1, B1 | Attempt at acceleration for \(0 \leq t \leq 2\); must be negative or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| \(58.5 = 12 \times 6 + 0.5 \times a \times 36\) so \(a = -0.75\) | M1, A1 | Use of *uvast* with 12 and 58.5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = -10 + \frac{9}{2}t - \frac{3}{8}t^2\) | M1, A1 | Differentiation |
| \(a(1) = -10 + \frac{9}{2} - \frac{3}{8} = -5.875\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(s = \int\left(12 - 10t + \frac{9}{4}t^2 - \frac{1}{8}t^3\right)dt\) | M1 | Attempt to integrate |
| \(= 12t - 5t^2 + \frac{3}{4}t^3 - \frac{1}{32}t^4 + C\) | A1, A1 | At least one term correct; All correct. Accept \(+ C\) omitted |
| \(s = 0\) when \(t = 0\) so \(C = 0\) | A1* | Clearly shown |
| \(s(8) = 32\) | A1 | cao (award even if A1* not given) |
| Answer | Marks | Guidance |
|---|---|---|
| \(s(2) = 9.5\) and \(s(4) = 8\) | B1 | Both calculated correctly from their \(s\). No further marks if their \(s(2) \leq s(4)\) |
| Displacement is negative; Car going backwards | E1, E1 | Do *not* need car going backwards *throughout* the interval |
| or Evaluate \(v(t)\) where \(2 < t < 4\) or appeal to shape of graph | B1 | e.g. \(v(3) = -1.125\). No further marks if their \(v \geq 0\) |
| Velocity is negative; Car going backwards | E1, E1 | Do *not* need car going backwards *throughout* the interval. [Award WW2 for 'car going backwards'; WW1 for velocity or displacement negative] |
# Question 6:
## Part (i)
| $0.5 \times 2 \times 12 + 0.5 \times 4 \times 12$ so $36$ m | M1, A1 | Attempt at sum of areas or equivalent. No extra areas. |
## Part (ii)
| $8 - \frac{36}{12} = 5$ seconds | B1 | cao |
## Part (iii)
| $-6$ m s$^{-2}$ | M1, B1 | Attempt at acceleration for $0 \leq t \leq 2$; must be negative or equivalent |
## Part (iv)
| $58.5 = 12 \times 6 + 0.5 \times a \times 36$ so $a = -0.75$ | M1, A1 | Use of *uvast* with 12 and 58.5 |
## Part (v)
| $a = -10 + \frac{9}{2}t - \frac{3}{8}t^2$ | M1, A1 | Differentiation |
| $a(1) = -10 + \frac{9}{2} - \frac{3}{8} = -5.875$ | A1 | cao |
## Part (vi)
| $s = \int\left(12 - 10t + \frac{9}{4}t^2 - \frac{1}{8}t^3\right)dt$ | M1 | Attempt to integrate |
| $= 12t - 5t^2 + \frac{3}{4}t^3 - \frac{1}{32}t^4 + C$ | A1, A1 | At least one term correct; All correct. Accept $+ C$ omitted |
| $s = 0$ when $t = 0$ so $C = 0$ | A1* | Clearly shown |
| $s(8) = 32$ | A1 | cao (award even if A1* not given) |
## Part (vii)
| $s(2) = 9.5$ and $s(4) = 8$ | B1 | Both calculated correctly from **their** $s$. No further marks if **their** $s(2) \leq s(4)$ |
| Displacement is negative; Car going backwards | E1, E1 | Do *not* need car going backwards *throughout* the interval |
| **or** Evaluate $v(t)$ where $2 < t < 4$ or appeal to shape of graph | B1 | e.g. $v(3) = -1.125$. No further marks if **their** $v \geq 0$ |
| Velocity is negative; Car going backwards | E1, E1 | Do *not* need car going backwards *throughout* the interval. [Award WW2 for 'car going backwards'; WW1 for velocity or displacement negative] |
---6 A toy car is travelling in a straight horizontal line.\\
One model of the motion for $0 \leqslant t \leqslant 8$, where $t$ is the time in seconds, is shown in the velocity-time graph Fig. 6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4957086c-fd1c-4cdc-bbdb-1959b3b21b2d-4_474_1196_580_424}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
(i) Calculate the distance travelled by the car from $t = 0$ to $t = 8$.\\
(ii) How much less time would the car have taken to travel this distance if it had maintained its initial speed throughout?\\
(iii) What is the acceleration of the car when $t = 1$ ?
From $t = 8$ to $t = 14$, the car travels 58.5 m with a new constant acceleration, $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(iv) Find $a$.
A second model for the velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, of the toy car is
$$v = 12 - 10 t + \frac { 9 } { 4 } t ^ { 2 } - \frac { 1 } { 8 } t ^ { 3 } , \text { for } 0 \leqslant t \leqslant 8$$
This model agrees with the values for $v$ given in Fig. 6 for $t = 0,2,4$ and 6. [Note that you are not required to verify this.] Use this second model to answer the following questions.\\
(v) Calculate the acceleration of the car when $t = 1$.\\
(vi) Initially the car is at A. Find an expression in terms of $t$ for the displacement of the car from A after the first $t$ seconds of its motion.
Hence find the displacement of the car from A when $t = 8$.\\
(vii) Explain with a reason what this model predicts for the motion of the car between $t = 2$ and $t = 4$.
\hfill \mbox{\textit{OCR MEI M1 2006 Q6 [18]}}