| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Velocity from acceleration by integration |
| Difficulty | Standard +0.3 This is a straightforward variable acceleration problem requiring integration with initial conditions. Students must integrate the given acceleration function to find velocity (part shown), solve a quadratic equation for time when v=0, then integrate velocity to find distance. While it involves non-constant acceleration (making it slightly above average), the algebraic manipulations are routine and the problem structure is standard for M1, with clear signposting through 'show that' parts. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = \pm\int -8 + 0.6t \; dt\) | M1 | Integrates accn or decn |
| \(v = \;^+/\text{-}(-8t + 0.6t^2/2) \; (+c)\) | A1 | Although only \(v = -8t + 0.6t^2/2 \; (+c)\) is correct |
| \(v = 32.5 - 8t + 0.3t^2\) | A1 [3] AG | ONLY FROM \(v = \int -8 + 0.6t \; dt\) OR \(v = -\int 8 - 0.6t \; dt\) and explicit \(t=0\), \(v = 32.5\) so \(c = 32.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.3t^2 - 8t + 32.5 = 0\) | M1 | Starts to solve 3 term QE, either the given answer in (i) or candidate's answer in (i) with \(v\) set \(= 0\). Needs valid formula or factors which give 2 correct coefficients |
| \(t = 5\) | A1 [2] | Accept as one of a pair only if the other value is \(65/3 = 21.66\ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s = \int 0.3t^2 - 8t + 32.5 \; dt\) | M1 | Integrates an expression for velocity |
| \(s = 0.3t^3/3 - 8t^2/2 + 32.5t \; (+c)\) | A1 | Accept omission of \(c\) |
| \(D = 0.3 \times 5^3/3 - 8 \times 5^2/2 + 32.5 \times 5 \; (+c)\) | M1 | Substitutes cv(smaller and \(+\)ve ans(ii)) or uses limits, \([\;]_0^{\text{smaller}+\text{vecv(ii)}}\) |
| \(D = 75\) | A1 [4] AG | Explicit evaluation needed. Accept \(+c\) |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = \pm\int -8 + 0.6t \; dt$ | M1 | Integrates accn or decn |
| $v = \;^+/\text{-}(-8t + 0.6t^2/2) \; (+c)$ | A1 | Although only $v = -8t + 0.6t^2/2 \; (+c)$ is correct |
| $v = 32.5 - 8t + 0.3t^2$ | A1 **[3]** AG | ONLY FROM $v = \int -8 + 0.6t \; dt$ OR $v = -\int 8 - 0.6t \; dt$ and explicit $t=0$, $v = 32.5$ so $c = 32.5$ |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.3t^2 - 8t + 32.5 = 0$ | M1 | Starts to solve 3 term QE, either the given answer in (i) or candidate's answer in (i) with $v$ set $= 0$. Needs valid formula or factors which give 2 correct coefficients |
| $t = 5$ | A1 **[2]** | Accept as one of a pair only if the other value is $65/3 = 21.66\ldots$ |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s = \int 0.3t^2 - 8t + 32.5 \; dt$ | M1 | Integrates an expression for velocity |
| $s = 0.3t^3/3 - 8t^2/2 + 32.5t \; (+c)$ | A1 | Accept omission of $c$ |
| $D = 0.3 \times 5^3/3 - 8 \times 5^2/2 + 32.5 \times 5 \; (+c)$ | M1 | Substitutes cv(smaller and $+$ve ans(ii)) or uses limits, $[\;]_0^{\text{smaller}+\text{vecv(ii)}}$ |
| $D = 75$ | A1 **[4]** AG | Explicit evaluation needed. Accept $+c$ |
---3 A car is travelling along a straight horizontal road with velocity $32.5 \mathrm {~ms} ^ { - 1 }$. The driver applies the brakes and the car decelerates at $( 8 - 0.6 t ) \mathrm { ms } ^ { - 2 }$, where $t \mathrm {~s}$ is the time which has elapsed since the brakes were first applied.\\
(i) Show that, while the car is decelerating, its velocity is $\left( 32.5 - 8 t + 0.3 t ^ { 2 } \right) \mathrm { m } \mathrm { s } ^ { - 1 }$.\\
(ii) Find the time taken to bring the car to rest.\\
(iii) Show that the distance travelled while the car is decelerating is 75 m .
\hfill \mbox{\textit{OCR M1 2012 Q3 [9]}}