| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring standard differentiation to find velocity and acceleration, then solving a quadratic equation. Part (i) involves routine calculus (v = dx/dt, a = dv/dt) and substitution. Part (ii) requires finding total distance and using average speed formula, which is a standard textbook exercise with no novel problem-solving required. Slightly easier than average due to the mechanical nature of the calculations. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Part (i) | ||
| \(v = 1.2t - 0.012t^2\) | M1 | For using \(v(t) = \dot{s}(t)\) |
| \([a(50) = 1.2 - 0.024 \times 50]\) | M1 | For using \(a(t) = \dot{v}(t)\) and evaluating \(a(50)\) |
| \(a = 0\) | A1 | AG |
| \(V = 30\) | B1 | 5 |
| Part (ii) | ||
| \([s = 0.6 \times 50^2 - 0.004 \times 50^3 (= 1000)]\) | B1 | |
| \(\left[\frac{1000 + s_2}{50 + t_2} = 27.5\right]\) | M1 | For using 'average speed = total distance /total time' |
| \([1000 + 30t_2 = 27.5(50 + t_2)]\) | M1 | For substituting \(s_2 = Vt_2\) and attempting to solve for \(t_2\) |
| \(t_2 = 150\) | A1 | |
| \(t = 200\) | A1 | 5 |
| (Alternative for part (ii)) | ||
| \(s_1 = 0.6 \times 50^2 - 0.004 \times 50^3 (= 1000)\) | B1 | |
| \(\left[\frac{(1000 + s_2)/t}{t = t - 50} = 27.5\right]\) | M1 | For using 'average speed = total distance /total time' with \(t_2 = t - 50\) (ft V and \(s_1\)) |
| \([27.5t = 1000 + 30(t - 50)]\) | M1 | For attempting to solve for \(t\) |
| \(t = 200\) | A1 | 5 |
| **Part (i)** | | |
|---|---|---|
| $v = 1.2t - 0.012t^2$ | M1 | For using $v(t) = \dot{s}(t)$ |
| $[a(50) = 1.2 - 0.024 \times 50]$ | M1 | For using $a(t) = \dot{v}(t)$ and evaluating $a(50)$ |
| $a = 0$ | A1 | AG |
| $V = 30$ | B1 | 5 |
| **Part (ii)** | | |
|---|---|---|
| $[s = 0.6 \times 50^2 - 0.004 \times 50^3 (= 1000)]$ | B1 | |
| $\left[\frac{1000 + s_2}{50 + t_2} = 27.5\right]$ | M1 | For using 'average speed = total distance /total time' |
| $[1000 + 30t_2 = 27.5(50 + t_2)]$ | M1 | For substituting $s_2 = Vt_2$ and attempting to solve for $t_2$ |
| $t_2 = 150$ | A1 | |
| $t = 200$ | A1 | 5 | ft 50 + $t_2$ (requires both M marks) |
| **(Alternative for part (ii))** | | |
|---|---|---|
| $s_1 = 0.6 \times 50^2 - 0.004 \times 50^3 (= 1000)$ | B1 | |
| $\left[\frac{(1000 + s_2)/t}{t = t - 50} = 27.5\right]$ | M1 | For using 'average speed = total distance /total time' with $t_2 = t - 50$ (ft V and $s_1$) |
| $[27.5t = 1000 + 30(t - 50)]$ | M1 | For attempting to solve for $t$ |
| $t = 200$ | A1 | 5 |7 A motorcyclist starts from rest at $A$ and travels in a straight line. For the first part of the motion, the motorcyclist's displacement $x$ metres from $A$ after $t$ seconds is given by $x = 0.6 t ^ { 2 } - 0.004 t ^ { 3 }$.\\
(i) Show that the motorcyclist's acceleration is zero when $t = 50$ and find the speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at this time.
For $t \geqslant 50$, the motorcyclist travels at constant speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) Find the value of $t$ for which the motorcyclist's average speed is $27.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\hfill \mbox{\textit{CAIE M1 2009 Q7 [10]}}