| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Standard +0.3 This is a straightforward kinematics question requiring differentiation of a given displacement function to find velocity and acceleration. All parts involve routine calculus techniques (finding when v=0 for part (i), finding maximum of v(t) for part (ii), evaluating derivatives at endpoints for part (iii)). The polynomial is given explicitly, making this slightly easier than average despite being multi-part. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([0.0000117(1200t^2 - 12t^3) = 0]\) | M1 | For differentiating and solving \(ds/dt = 0\) |
| \(1200t^2 = 12t^3 \Rightarrow t = 0,\ 100\) | A1 | Accept just \(t = 100\), if used to find distance AB |
| Distance \(AB = 1170\) m | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For differentiating again and solving \(d^2s/dt^2 = 0\) | |
| \(2400t - 36t^2 = 0 \Rightarrow t = 0,\ 200/3\) | A1 | Accept just \(t = 200/3\), if used to find \(v_{max}\) |
| \([v_{max} = 0.0000117\{1200(200/3)^2 - 12(200/3)^3\}]\) | M1 | For substituting into \(v(t)\) |
| Maximum speed is \(20.8\ ms^{-1}\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At A \(a(t) = 0\) | B1 | |
| At B \(a(t) = 0.0000117(2400 \times 100 - 36 \times 100^2) = -1.40\ ms^{-2}\ (-1.404\) exact\()\) | B1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sketch has \(v\) increasing from 0 to maximum and decreasing to 0, with maximum closer to \(t = 100\) than \(t = 0\) | B1 | |
| Sketch has zero gradient at \(t = 0\) and inflexion closer to \(t = 0\) than \(t = 100\) | B1 | [2] |
## Question 7:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[0.0000117(1200t^2 - 12t^3) = 0]$ | M1 | For differentiating and solving $ds/dt = 0$ |
| $1200t^2 = 12t^3 \Rightarrow t = 0,\ 100$ | A1 | Accept just $t = 100$, if used to find distance AB |
| Distance $AB = 1170$ m | A1 | [3] |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For differentiating again and solving $d^2s/dt^2 = 0$ |
| $2400t - 36t^2 = 0 \Rightarrow t = 0,\ 200/3$ | A1 | Accept just $t = 200/3$, if used to find $v_{max}$ |
| $[v_{max} = 0.0000117\{1200(200/3)^2 - 12(200/3)^3\}]$ | M1 | For substituting into $v(t)$ |
| Maximum speed is $20.8\ ms^{-1}$ | A1 | [4] |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| At A $a(t) = 0$ | B1 | |
| At B $a(t) = 0.0000117(2400 \times 100 - 36 \times 100^2) = -1.40\ ms^{-2}\ (-1.404$ exact$)$ | B1 | [2] |
### Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sketch has $v$ increasing from 0 to maximum and decreasing to 0, with maximum closer to $t = 100$ than $t = 0$ | B1 | |
| Sketch has zero gradient at $t = 0$ and inflexion closer to $t = 0$ than $t = 100$ | B1 | [2] |7 A car driver makes a journey in a straight line from $A$ to $B$, starting from rest. The speed of the car increases to a maximum, then decreases until the car is at rest at $B$. The distance travelled by the car $t$ seconds after leaving $A$ is $0.0000117 \left( 400 t ^ { 3 } - 3 t ^ { 4 } \right)$ metres.\\
(i) Find the distance $A B$.\\
(ii) Find the maximum speed of the car.\\
(iii) Find the acceleration of the car
\begin{enumerate}[label=(\alph*)]
\item as it starts from $A$,
\item as it arrives at $B$.\\
(iv) Sketch the velocity-time graph for the journey.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2013 Q7 [11]}}