| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find position by integrating velocity |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics question involving differentiation, integration, and 2D kinematics. Parts (i)-(iv) require routine calculus and solving quadratic equations with no novel insight. Part (v) is a plotting exercise. Slightly easier than average due to straightforward algebraic manipulation and clear problem structure. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10h Vectors in kinematics: uniform acceleration in vector form3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v_x = 8 - 4t\) | M1 | Either differentiating, or finding \(u\) and \(a\) from \(x\) and use of \(v = u + at\) |
| \(v_x = 0 \Leftrightarrow t = 2\), so at \(t = 2\) | A1, F1 | FT their \(v_x = 0\) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \int(3t^2 - 8t + 4)\ \text{d}t\) | M1 | Integrating \(v_y\) with at least one correct integrated term |
| \(= t^3 - 4t^2 + 4t + c\) | A1 | All correct. Accept no arbitrary constant |
| \(y=3\) when \(t=1\) so \(3 = 1 - 4 + 4 + c\) | M1 | Clear evidence |
| \(c = 3-1 = 2\) and \(y = t^3 - 4t^2 + 4t + 2\) | E1 | Clearly shown and stated |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| We need \(x=0\) so \(8t - 2t^2 = 0\), \(t=0\) or \(t=4\) | M1, A1 | May be implied. Must have both |
| \(t=0\) gives \(y=2\) so \(2\) m | A1 | Condone \(2\mathbf{j}\) |
| \(t=4\) gives \(y = 4^3 - 4^3 + 16 + 2 = 18\) so \(18\) m | A1 | Condone \(18\mathbf{j}\) |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| We need \(v_x = v_y = 0\) | M1 | Either recognises \(v_x = 0\) when \(t=2\), or finds time(s) when \(v_y=0\), or states/implies \(v_x = v_y = 0\) |
| From above, \(v_x = 0\) only when \(t=2\); evaluate \(v_y(2)\) | M1 | Considers \(v_x = 0\) and \(v_y = 0\) with their time(s) |
| \(v_y(2) = 0\) [\(t-2\) is a factor], so yes only at \(t=2\) | A1 | \(t=2\) recognised as only value (accept as evidence only; \(t=2\) used below). For last 2 marks, no credit lost for reference to \(t=\frac{2}{3}\) |
| At \(t=2\), position is \((8, 2)\) | B1 | May be implied |
| Distance is \(\sqrt{8^2 + 2^2} = \sqrt{68}\) m (\(8.25\) 3 s.f.) | B1 | FT from their position. Accept one position followed through correctly |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=0, 1\) give \((0,2)\) and \((6,3)\) | B1 | At least one value \(0\leq t < 2\) correctly calculated. Need not be plotted |
| B1 | Must be \(x\)-\(y\) curve. Accept sketch. Ignore curve outside interval for \(t\). Accept unlabelled axes. Condone use of line segments | |
| B1 | At least three correct points used in \(x\)-\(y\) graph or sketch. General shape correct. Do not condone use of line segments | |
| 3 |
# Question 8:
**(i)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v_x = 8 - 4t$ | M1 | Either differentiating, or finding $u$ and $a$ from $x$ and use of $v = u + at$ |
| $v_x = 0 \Leftrightarrow t = 2$, so at $t = 2$ | A1, F1 | FT their $v_x = 0$ |
| | **3** | |
**(ii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \int(3t^2 - 8t + 4)\ \text{d}t$ | M1 | Integrating $v_y$ with at least one correct integrated term |
| $= t^3 - 4t^2 + 4t + c$ | A1 | All correct. Accept no arbitrary constant |
| $y=3$ when $t=1$ so $3 = 1 - 4 + 4 + c$ | M1 | Clear evidence |
| $c = 3-1 = 2$ and $y = t^3 - 4t^2 + 4t + 2$ | E1 | Clearly shown and stated |
| | **4** | |
**(iii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| We need $x=0$ so $8t - 2t^2 = 0$, $t=0$ or $t=4$ | M1, A1 | May be implied. Must have both |
| $t=0$ gives $y=2$ so $2$ m | A1 | Condone $2\mathbf{j}$ |
| $t=4$ gives $y = 4^3 - 4^3 + 16 + 2 = 18$ so $18$ m | A1 | Condone $18\mathbf{j}$ |
| | **4** | |
**(iv)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| We need $v_x = v_y = 0$ | M1 | Either recognises $v_x = 0$ when $t=2$, or finds time(s) when $v_y=0$, or states/implies $v_x = v_y = 0$ |
| From above, $v_x = 0$ only when $t=2$; evaluate $v_y(2)$ | M1 | Considers $v_x = 0$ **and** $v_y = 0$ with their time(s) |
| $v_y(2) = 0$ [$t-2$ is a factor], so yes only at $t=2$ | A1 | $t=2$ recognised as only value (accept as evidence only; $t=2$ used below). For last 2 marks, no credit lost for reference to $t=\frac{2}{3}$ |
| At $t=2$, position is $(8, 2)$ | B1 | May be implied |
| Distance is $\sqrt{8^2 + 2^2} = \sqrt{68}$ m ($8.25$ 3 s.f.) | B1 | FT from their position. Accept one position followed through correctly |
| | **5** | |
**(v)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=0, 1$ give $(0,2)$ and $(6,3)$ | B1 | At least one value $0\leq t < 2$ correctly calculated. Need not be plotted |
| | B1 | Must be $x$-$y$ curve. Accept sketch. Ignore curve outside interval for $t$. Accept unlabelled axes. Condone use of line segments |
| | B1 | At least three correct points **used** in $x$-$y$ graph or sketch. General shape correct. Do not condone use of line segments |
| | **3** | |8 A toy boat moves in a horizontal plane with position vector $\mathbf { r } = x \mathbf { i } + y \mathbf { j }$, where $\mathbf { i }$ and $\mathbf { j }$ are the standard unit vectors east and north respectively. The origin of the position vectors is at O . The displacements $x$ and $y$ are in metres.
First consider only the motion of the boat parallel to the $x$-axis. For this motion
$$x = 8 t - 2 t ^ { 2 }$$
The velocity of the boat in the $x$-direction is $v _ { x } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Find an expression in terms of $t$ for $v _ { x }$ and determine when the boat instantaneously has zero speed in the $x$-direction.
Now consider only the motion of the boat parallel to the $y$-axis. For this motion
$$v _ { y } = ( t - 2 ) ( 3 t - 2 )$$
where $v _ { y } \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the velocity of the boat in the $y$-direction at time $t$ seconds.\\
(ii) Given that $y = 3$ when $t = 1$, use integration to show that $y = t ^ { 3 } - 4 t ^ { 2 } + 4 t + 2$.
The position vector of the boat is given in terms of $t$ by $\mathbf { r } = \left( 8 t - 2 t ^ { 2 } \right) \mathbf { i } + \left( t ^ { 3 } - 4 t ^ { 2 } + 4 t + 2 \right) \mathbf { j }$.\\
(iii) Find the time(s) when the boat is due north of O and also the distance of the boat from O at any such times.\\
(iv) Find the time(s) when the boat is instantaneously at rest. Find the distance of the boat from O at any such times.\\
(v) Plot a graph of the path of the boat for $0 \leqslant t \leqslant 2$.
\hfill \mbox{\textit{OCR MEI M1 2009 Q8 [19]}}