| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Moderate -0.3 This is a straightforward calculus application question requiring differentiation of a given displacement function to find velocity, then finding the maximum by setting the derivative to zero. Part (a) involves simple substitution to find when velocity equals zero, and part (b) is a standard optimization problem with no conceptual challenges—slightly easier than average due to the explicit formula provided and routine techniques required. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For differentiation of \(s\) | *M1 | |
| \(v = 0.004(150t - 3t^2)\ [= 0.6t - 0.012t^2]\) | A1 | |
| \(v = 0\) when \(t = 50\). At \(t = 50\), \(s = 0.004(75 \times 50^2 - 50^3) = 0.3 \times 50^2 - 0.004 \times 50^3\) | DM1 | Solve \(v = 0\) for \(t\) and substitute this value into \(s\). |
| Distance \(AB = 250\) m | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to determine stationary points for \(v\) by differentiation or by use of symmetry \([a = 0.004(150 - 6t) = 0.6 - 0.024t]\) or using symmetry attempt to find the mid-point between \(t = 0\) and their \(t\) value at \(v = 0\) | *M1 | If symmetry used then an attempt to find the required mid-point must be seen. |
| Maximum \(v\) when \(a = 0\) so \(t = 25\); or finding the mid-point if symmetry is used e.g. \(v = 0.004(150 \times 25 - 3 \times 25^2) = 0.6 \times 25 - 0.012 \times 25^2\ [= 7.5\ \text{ms}^{-1}]\) | DM1 | Attempt to solve \(a = 0\) or use symmetry to find the relevant \(t\) value. |
| Maximum velocity \(= 7.5\) ms\(^{-1}\) | A1 | |
| Alternative: Attempt to velocity as \(v = -0.012\left[(t-25)^2 - 25^2\right]\) | M1* | Attempt to complete the square for their velocity as far as \(k\left[(t-a)^2 - a^2\right]\) |
| \(v = -0.012(t-25)^2 + 0.012 \times 25^2\) and select \(t = 25\) as the maximum point. | DM1 | Or select the \(0.012 \times 25^2\) term as the maximum velocity. |
| Maximum \(= [0.012 \times 625 =]\ 7.5\) ms\(^{-1}\) | A1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For differentiation of $s$ | *M1 | |
| $v = 0.004(150t - 3t^2)\ [= 0.6t - 0.012t^2]$ | A1 | |
| $v = 0$ when $t = 50$. At $t = 50$, $s = 0.004(75 \times 50^2 - 50^3) = 0.3 \times 50^2 - 0.004 \times 50^3$ | DM1 | Solve $v = 0$ for $t$ and substitute this value into $s$. |
| Distance $AB = 250$ m | A1 | AG |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to determine stationary points for $v$ by differentiation or by use of symmetry $[a = 0.004(150 - 6t) = 0.6 - 0.024t]$ or using symmetry attempt to find the mid-point between $t = 0$ and their $t$ value at $v = 0$ | *M1 | If symmetry used then an attempt to find the required mid-point must be seen. |
| Maximum $v$ when $a = 0$ so $t = 25$; or finding the mid-point if symmetry is used e.g. $v = 0.004(150 \times 25 - 3 \times 25^2) = 0.6 \times 25 - 0.012 \times 25^2\ [= 7.5\ \text{ms}^{-1}]$ | DM1 | Attempt to solve $a = 0$ or use symmetry to find the relevant $t$ value. |
| Maximum velocity $= 7.5$ ms$^{-1}$ | A1 | |
| **Alternative:** Attempt to velocity as $v = -0.012\left[(t-25)^2 - 25^2\right]$ | M1* | Attempt to complete the square for their velocity as far as $k\left[(t-a)^2 - a^2\right]$ |
| $v = -0.012(t-25)^2 + 0.012 \times 25^2$ and select $t = 25$ as the maximum point. | DM1 | Or select the $0.012 \times 25^2$ term as the maximum velocity. |
| Maximum $= [0.012 \times 625 =]\ 7.5$ ms$^{-1}$ | A1 | |
---4 A cyclist starts from rest at a point $A$ and travels along a straight road $A B$, coming to rest at $B$. The displacement of the cyclist from $A$ at time $t \mathrm {~s}$ after the start is $s \mathrm {~m}$, where
$$s = 0.004 \left( 75 t ^ { 2 } - t ^ { 3 } \right)$$
\begin{enumerate}[label=(\alph*)]
\item Show that the distance $A B$ is 250 m .
\item Find the maximum velocity of the cyclist.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q4 [7]}}