| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Velocity from acceleration by integration |
| Difficulty | Standard +0.3 This is a standard M1 variable acceleration question requiring integration of a quadratic to find velocity, applying initial conditions, and analyzing motion direction. The steps are routine: (i) solve a quadratic equation, (ii) integrate and use given condition to find constant, (iii) show velocity changes sign. While it requires multiple techniques, all are standard M1 procedures with no novel insight needed, making it slightly easier than average A-level difficulty. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((t-3)(t-6) = 0\) | M1 | Solve 3 term QE, 2 correct coefficients if factorising, or using formula \(9+/-\sqrt{9}/2\) |
| \(t = 3, 6\) | A1 | "By inspection" both values M1A1, one value M0A0 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = \int(t^2 - 9t + 18)\,dt\) | M1* | Attempts integration of \(a(t)dt\), maximum one wrong term |
| \(v = t^3/3 - 9t^2/2 + 18t\ (+c)\) | A1 | Accept omission of \(+c\) |
| \(3^3/3 - 9 \times 3^2/2 + 18 \times 3 + c = 9\) | D*M1 | Uses \(v(3) = 9\) |
| \((v =) -13.5\) m s\(^{-1}\) | A1 | Must be negative, and goes beyond \(c = -13.5\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v(1) = 1/3 - 9/2 + 18 - 13.5 = 0.333\) | M1 | Finds \(v(1)\ (=1/3)\) |
| Changed sign so direction of motion has changed | A1 | Accurate values: \(v(0) = -13.5\), \(v(0.5) = -5.58\), \(v(0.9) = -0.702\) |
| [2] |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(t-3)(t-6) = 0$ | M1 | Solve 3 term QE, 2 correct coefficients if factorising, or using formula $9+/-\sqrt{9}/2$ |
| $t = 3, 6$ | A1 | "By inspection" both values M1A1, one value M0A0 |
| **[2]** | | |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = \int(t^2 - 9t + 18)\,dt$ | M1* | Attempts integration of $a(t)dt$, maximum one wrong term |
| $v = t^3/3 - 9t^2/2 + 18t\ (+c)$ | A1 | Accept omission of $+c$ |
| $3^3/3 - 9 \times 3^2/2 + 18 \times 3 + c = 9$ | D*M1 | Uses $v(3) = 9$ |
| $(v =) -13.5$ m s$^{-1}$ | A1 | Must be negative, and goes beyond $c = -13.5$ |
| **[4]** | | |
## Question 4(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v(1) = 1/3 - 9/2 + 18 - 13.5 = 0.333$ | M1 | Finds $v(1)\ (=1/3)$ |
| Changed sign so direction of motion has changed | A1 | Accurate values: $v(0) = -13.5$, $v(0.5) = -5.58$, $v(0.9) = -0.702$ |
| **[2]** | | |4 The acceleration of a particle $P$ moving in a straight line is $\left( t ^ { 2 } - 9 t + 18 \right) \mathrm { ms } ^ { - 2 }$, where $t$ is the time in seconds.\\
(i) Find the values of $t$ for which the acceleration is zero.\\
(ii) It is given that when $t = 3$ the velocity of $P$ is $9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the velocity of $P$ when $t = 0$.\\
(iii) Show that the direction of motion of $P$ changes before $t = 1$.
\hfill \mbox{\textit{OCR M1 2013 Q4 [8]}}