| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Displacement from velocity by integration |
| Difficulty | Standard +0.3 This is a straightforward calculus-based mechanics problem requiring differentiation to find acceleration, using the minimum condition (dv/dt = 0) to find k, and integration to find distance. All steps are standard M1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
6 A swimmer $C$ swims with velocity $v \mathrm {~ms} ^ { - 1 }$ in a swimming pool. At time $t \mathrm {~s}$ after starting, $v = 0.006 t ^ { 2 } - 0.18 t + k$, where $k$ is a constant. $C$ swims from one end of the pool to the other in 28.4 s .\\
(i) Find the acceleration of $C$ in terms of $t$.\\
(ii) Given that the minimum speed of $C$ is $0.65 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, show that $k = 2$.\\
(iii) Express the distance travelled by $C$ in terms of $t$, and calculate the length of the pool.
\hfill \mbox{\textit{OCR M1 2010 Q6 [12]}}