| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding constants from motion conditions |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring integration of a velocity function to find displacement, then solving for a constant. Part (b) involves basic calculus (finding stationary points). The techniques are standard M1 content with no conceptual challenges—slightly easier than average due to the direct application of routine methods. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For attempt at integration | M1* | The power of \(t\) must increase by 1 with a change of coefficient in the same term. Use of \(s = vt\) scores M0. |
| \(\dfrac{1}{2+1}kt^{2+1} - \dfrac{4}{2}t^{1+1} + 3t \left[= \dfrac{1}{3}kt^3 - 2t^2 + 3t\right] [+c]\) | A1 | Allow unsimplified. |
| \(\dfrac{1}{3}k\times2^3 - 2\times2^2 + 3\times2\ [-0] = 6\) | DM1 | Use of limits 0 and 2 with 6 to form an equation in \(k\) only (without \(c\) but allow with \(+c - c\)). |
| \(k = 3\) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2\times3t - 4\); or at min value \(t = \dfrac{-b}{2a} = \dfrac{4}{2\times3}\) | M1 | For attempt at differentiation. Must have expression of the form \(at + b\) with \(a \neq 3\), unless their \(k = \frac{3}{2}\). Allow \(2kt - 4\). |
| \(\left[2\times3t - 4 = 0 \Rightarrow\right]\ t = \dfrac{2}{3}\) | A1FT | OE. FT their \(k\): \(t = \dfrac{2}{\text{their } k}\). Allow without working. |
| \(v\left[= 3\times\left(\dfrac{2}{3}\right)^2 - 4\times\dfrac{2}{3} + 3\right] = \dfrac{5}{3}\ \text{ms}^{-1}\) | A1 | OE. Allow 1.67 or better for \(v\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt at completing the square | (M1) | Must have \(\left(t - \dfrac{2}{3}\right)^2\) OE, or \(\left(t - \dfrac{2}{\text{their }k}\right)^2\). |
| \(3\!\left(t - \dfrac{2}{3}\right)^2 - \dfrac{4}{3} + 3\) | (A1FT) | FT their \(k\): \(k\!\left(t-\dfrac{2}{k}\right)^2 - \dfrac{4}{k} + 3\). |
| \(v = \dfrac{5}{3}\ \text{ms}^{-1}\) | (A1) | OE. Allow 1.67 or better. |
| 3 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For attempt at integration | M1* | The power of $t$ must increase by 1 with a change of coefficient in the same term. Use of $s = vt$ scores M0. |
| $\dfrac{1}{2+1}kt^{2+1} - \dfrac{4}{2}t^{1+1} + 3t \left[= \dfrac{1}{3}kt^3 - 2t^2 + 3t\right] [+c]$ | A1 | Allow unsimplified. |
| $\dfrac{1}{3}k\times2^3 - 2\times2^2 + 3\times2\ [-0] = 6$ | DM1 | Use of limits 0 and 2 with 6 to form an equation in $k$ only (without $c$ but allow with $+c - c$). |
| $k = 3$ | A1 | |
| | **4** | |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\times3t - 4$; or at min value $t = \dfrac{-b}{2a} = \dfrac{4}{2\times3}$ | M1 | For attempt at differentiation. Must have expression of the form $at + b$ with $a \neq 3$, unless their $k = \frac{3}{2}$. Allow $2kt - 4$. |
| $\left[2\times3t - 4 = 0 \Rightarrow\right]\ t = \dfrac{2}{3}$ | A1FT | OE. FT their $k$: $t = \dfrac{2}{\text{their } k}$. Allow without working. |
| $v\left[= 3\times\left(\dfrac{2}{3}\right)^2 - 4\times\dfrac{2}{3} + 3\right] = \dfrac{5}{3}\ \text{ms}^{-1}$ | A1 | OE. Allow 1.67 or better for $v$. |
**Alternative: Completing the square**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt at completing the square | (M1) | Must have $\left(t - \dfrac{2}{3}\right)^2$ OE, or $\left(t - \dfrac{2}{\text{their }k}\right)^2$. |
| $3\!\left(t - \dfrac{2}{3}\right)^2 - \dfrac{4}{3} + 3$ | (A1FT) | FT their $k$: $k\!\left(t-\dfrac{2}{k}\right)^2 - \dfrac{4}{k} + 3$. |
| $v = \dfrac{5}{3}\ \text{ms}^{-1}$ | (A1) | OE. Allow 1.67 or better. |
| | **3** | |4 A particle travels in a straight line. The velocity of the particle at time $t \mathrm {~s}$ after leaving a point $O$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where
$$v = k t ^ { 2 } - 4 t + 3$$
The distance travelled by the particle in the first 2 s of its motion is 6 m . You may assume that $v > 0$ in the first 2s of its motion.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$.
\item Find the value of the minimum velocity of the particle. You do not need to show that this velocity is a minimum.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2024 Q4 [7]}}