| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding constants from motion conditions |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring differentiation to find acceleration, solving for a constant, then integration for distance. All steps are standard A-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = k + 0.06t\) | B1 | |
| \(k + 0.06(20) = 1.3\) | M1 | Use of \(t = 20\) and \(a = 1.3\) in their \(a\) |
| \(k = 1.3 - 1.2 = 0.1\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(s = 0.05t^2 + 0.01t^3\ (+c)\) | M1*, A1ft | Attempt to integrate – all powers increased by 1 |
| \(t = 0, s = 0 \Rightarrow c = 0\) | B1 | From a correct expression for \(s\) |
| \(t = 20, v = 14\) | B1ft | \(12 + 20k\) |
| \(s_1 = 0.05(20)^2 + 0.01(20)^3\) | dep*M1 | Finding distance travelled after 20s (for reference \(s_1 = 100\)) |
| \(25^2 = 14^2 + 2(1.3)s_2\) | M1 | Use of \(v^2 = u^2 + 2as\) with \(v = 25\) and \(a = 1.3\) and their \(u\) |
| Total distance \(= s_1 + s_2 = 265\text{m}\) | A1 [7] | All previous marks must have been awarded |
## Question 11:
### Part (i):
$a = k + 0.06t$ | B1 | |
$k + 0.06(20) = 1.3$ | M1 | Use of $t = 20$ and $a = 1.3$ in their $a$ |
$k = 1.3 - 1.2 = 0.1$ | A1 [3] | |
### Part (ii):
$s = 0.05t^2 + 0.01t^3\ (+c)$ | M1*, A1ft | Attempt to integrate – all powers increased by 1 |
$t = 0, s = 0 \Rightarrow c = 0$ | B1 | From a correct expression for $s$ |
$t = 20, v = 14$ | B1ft | $12 + 20k$ |
$s_1 = 0.05(20)^2 + 0.01(20)^3$ | dep*M1 | Finding distance travelled after 20s (for reference $s_1 = 100$) |
$25^2 = 14^2 + 2(1.3)s_2$ | M1 | Use of $v^2 = u^2 + 2as$ with $v = 25$ and $a = 1.3$ and their $u$ |
Total distance $= s_1 + s_2 = 265\text{m}$ | A1 [7] | All previous marks must have been awarded |
---11 The velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ of a car at time $t \mathrm {~s}$, during the first 20 s of its journey, is given by $v = k t + 0.03 t ^ { 2 }$, where $k$ is a constant. When $t = 20$ the acceleration of the car is $1.3 \mathrm {~ms} ^ { - 2 }$. For $t > 20$ the car continues its journey with constant acceleration $1.3 \mathrm {~ms} ^ { - 2 }$ until its speed reaches $25 \mathrm {~ms} ^ { - 1 }$.\\
(i) Find the value of $k$.\\
(ii) Find the total distance the car has travelled when its speed reaches $25 \mathrm {~ms} ^ { - 1 }$.
\hfill \mbox{\textit{OCR H240/03 2018 Q11 [10]}}