| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Moderate -0.3 This is a standard variable acceleration problem requiring integration with initial conditions and straightforward kinematics. Part (i) is a 'show that' requiring one integration (v = ∫a dt), part (ii) requires another integration (s = ∫v dt) plus substitution of values, and part (iii) requires equating distances. All steps are routine M1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(v = \int(4-t)\,dt\) | M1 | Attempt to integrate |
| \(v = 4t - \frac{1}{2}t^2 + c \quad (t=0, v=0 \Rightarrow c=0)\) | Condone no mention of arbitrary constant | |
| \(v = 4t - \frac{1}{2}t^2\) for \(0 \leq t \leq 4\) | A1 | |
| When \(t=4\), \(v=8\) and for \(t>4\), \(a=0\) so \(v=8\) for \(t>4\) | B1 | \(a=0\) must be seen or implied |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(s = \int(4t - \frac{1}{2}t^2)\,dt\) | M1 | |
| \(s = 2t^2 - \frac{1}{6}t^3\) | A1 | Again condone no mention of arbitrary constant |
| When \(t=4\), Nina has travelled \(2\times4^2 - \frac{1}{6}\times4^3 = 21\frac{1}{3}\) m | A1 | |
| When \(t = 5\frac{1}{3}\), Nina has travelled \(21\frac{1}{3} + 8\times1\frac{1}{3} = 32\) m | F1 | Allow follow through from their \(21\frac{1}{3}\). Exact answer required; if rounded to 32, award 0 |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| When \(t = 5\frac{1}{3}\), Marie has run \(6\times5\frac{1}{3} = 32\) m. Nina has also run 32 m so caught up Marie | B1 | Allow equivalent argument. This mark is dependent on answer 32 in part (ii) but allow where it is a rounded answer and rounding can be in part (iii) |
| [1] |
## Question 3(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $v = \int(4-t)\,dt$ | M1 | Attempt to integrate |
| $v = 4t - \frac{1}{2}t^2 + c \quad (t=0, v=0 \Rightarrow c=0)$ | | Condone no mention of arbitrary constant |
| $v = 4t - \frac{1}{2}t^2$ for $0 \leq t \leq 4$ | A1 | |
| When $t=4$, $v=8$ and for $t>4$, $a=0$ so $v=8$ for $t>4$ | B1 | $a=0$ must be seen or implied |
| | **[3]** | |
---
## Question 3(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $s = \int(4t - \frac{1}{2}t^2)\,dt$ | M1 | |
| $s = 2t^2 - \frac{1}{6}t^3$ | A1 | Again condone no mention of arbitrary constant |
| When $t=4$, Nina has travelled $2\times4^2 - \frac{1}{6}\times4^3 = 21\frac{1}{3}$ m | A1 | |
| When $t = 5\frac{1}{3}$, Nina has travelled $21\frac{1}{3} + 8\times1\frac{1}{3} = 32$ m | F1 | Allow follow through from their $21\frac{1}{3}$. Exact answer required; if rounded to 32, award 0 |
| | **[4]** | |
---
## Question 3(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| When $t = 5\frac{1}{3}$, Marie has run $6\times5\frac{1}{3} = 32$ m. Nina has also run 32 m so caught up Marie | B1 | Allow equivalent argument. This mark is dependent on answer 32 in part (ii) but allow where it is a rounded answer and rounding can be in part (iii) |
| | **[1]** | |
---3 Two girls, Marie and Nina, are members of an Olympic hockey team. They are doing fitness training. Marie runs along a straight line at a constant speed of $6 \mathrm {~ms} ^ { - 1 }$.
Nina is stationary at a point O on the line until Marie passes her. Nina immediately runs after Marie until she catches up with her.
The time, $t \mathrm {~s}$, is measured from the moment when Nina starts running. So when $t = 0$, both girls are at O .\\
Nina's acceleration, $a \mathrm {~ms} ^ { - 2 }$, is given by
$$\begin{array} { l l }
a = 4 - t & \text { for } 0 \leqslant t \leqslant 4 , \\
a = 0 & \text { for } t > 4 .
\end{array}$$
(i) Show that Nina's speed, $v \mathrm {~ms} ^ { - 1 }$, is given by
$$\begin{array} { l l }
v = 4 t - \frac { 1 } { 2 } t ^ { 2 } & \text { for } 0 \leqslant t \leqslant 4 , \\
v = 8 & \text { for } t > 4 .
\end{array}$$
(ii) Find an expression for the distance Nina has run at time $t$, for $0 \leqslant t \leqslant 4$.
Find how far Nina has run when $t = 4$ and when $t = 5 \frac { 1 } { 3 }$.\\
(iii) Show that Nina catches up with Marie when $t = 5 \frac { 1 } { 3 }$.
\hfill \mbox{\textit{OCR MEI M1 2012 Q3 [8]}}