| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring standard calculus techniques: differentiation for acceleration, integration for displacement, and finding maximum velocity by setting dv/dt = 0. Part (i) involves routine verification and substitution, while part (ii) requires finding the maximum velocity first then solving a quadratic. All steps are standard textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s = 0.3t^2 - 0.01t^3\) | M1 | For integration |
| \(s(5) = 0.3 \times 5^2 - 0.01 \times 5^3 = 6.25\) | A1 | |
| \(a = 0.6 - 0.06t\) | M1 | For differentiation |
| \(a(5) = 0.6 - 0.06 \times 5 = 0.3\ \text{ms}^{-2}\) | A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Maximum velocity is when \(0.6 - 0.06t = 0\) | M1 | For setting \(a = 0\) |
| \([t = 10]\) | M1 | For solving \(a = 0\) |
| Max velocity \(= 3\ \text{ms}^{-1}\) | A1 | |
| \(0.6t - 0.03t^2 = 1.5\) | Setting velocity \(=\) half its maximum and attempting to solve a three term quadratic | |
| \([t^2 - 20t + 50 = 0]\) | M1 | |
| Times are \(2.93\) s | A1 | |
| and \(17.07\) s | A1 (6) |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = 0.3t^2 - 0.01t^3$ | M1 | For integration |
| $s(5) = 0.3 \times 5^2 - 0.01 \times 5^3 = 6.25$ | A1 | |
| $a = 0.6 - 0.06t$ | M1 | For differentiation |
| $a(5) = 0.6 - 0.06 \times 5 = 0.3\ \text{ms}^{-2}$ | A1 (4) | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Maximum velocity is when $0.6 - 0.06t = 0$ | M1 | For setting $a = 0$ |
| $[t = 10]$ | M1 | For solving $a = 0$ |
| Max velocity $= 3\ \text{ms}^{-1}$ | A1 | |
| $0.6t - 0.03t^2 = 1.5$ | | Setting velocity $=$ half its maximum and attempting to solve a three term quadratic |
| $[t^2 - 20t + 50 = 0]$ | M1 | |
| Times are $2.93$ s | A1 | |
| and $17.07$ s | A1 (6) | |6 A particle $P$ moves in a straight line, starting from a point $O$. The velocity of $P$, measured in $\mathrm { m } \mathrm { s } ^ { - 1 }$, at time $t \mathrm {~s}$ after leaving $O$ is given by
$$v = 0.6 t - 0.03 t ^ { 2 }$$
(i) Verify that, when $t = 5$, the particle is 6.25 m from $O$. Find the acceleration of the particle at this time.\\
(ii) Find the values of $t$ at which the particle is travelling at half of its maximum velocity.
\hfill \mbox{\textit{CAIE M1 2015 Q6 [10]}}