| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Standard +0.3 This is a straightforward calculus-based mechanics question requiring differentiation of a polynomial to find velocity and acceleration, then finding when velocity is minimum (by setting acceleration to zero) and when direction changes (velocity equals zero). All steps are standard M1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.18(t^2 - 5t - 4/3)\) gives \((t-2.5)^2\) | M1 | By completing the square |
| \([-91/12]\) hence \(t = 2.5\) | A1 | Can return to formula for \(v(t)\) or calculate \( |
## Question 6(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.18(t^2 - 5t - 4/3)$ gives $(t-2.5)^2$ | M1 | By completing the square |
| $[-91/12]$ hence $t = 2.5$ | A1 | Can return to formula for $v(t)$ or calculate $|0.18x - 91/12|$ |
---6 A particle $P$ moves in a straight line. At time $t$ s after passing through a point $O$ of the line, the displacement of $P$ from $O$ is $x \mathrm {~m}$. Given that $x = 0.06 t ^ { 3 } - 0.45 t ^ { 2 } - 0.24 t$, find\\
(i) the velocity and the acceleration of $P$ when $t = 0$,\\
(ii) the value of $x$ when $P$ has its minimum velocity, and the speed of $P$ at this instant,\\
(iii) the positive value of $t$ when the direction of motion of $P$ changes.
\hfill \mbox{\textit{OCR M1 2013 Q6 [14]}}