| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding when particle at rest |
| Difficulty | Moderate -0.8 This is a straightforward kinematics question with non-constant acceleration. Part (a) requires solving a simple quadratic equation (v=0), part (b) is direct differentiation at t=0, part (c) uses completing the square or calculus to find a minimum, and part (d) involves integration with sign consideration. All parts use standard M1 techniques with no novel problem-solving required, making it easier than average. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4t^2 - 20t + 21 = (2t-3)(2t-7) = 0 \rightarrow t = \ldots\) | M1 | For setting \(v = 0\) and attempting to solve \(v = 0\) |
| \(t = 1.5\) and \(t = 3.5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = 8t - 20\), \(a(0) = \ldots\) | M1 | For using \(a = dv/dt\) and evaluating for \(t = 0\) |
| \(a = -20\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(8t - 20 = 0\), \(t = 2.5 \rightarrow v = \ldots\) or \(v = (2t-5)^2 - 4\), \(v_{\min} = \ldots\) | M1 | For setting \(a = 0\), attempting to solve for \(t\) and substituting to obtain \(v\), or for attempting to complete the square on the expression for \(v\) |
| \(v_{\min} = -4 \text{ ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s = \int(4t^2 - 20t + 21)\, dt\) | M1 | For using \(s = \int v\, dt\) and attempting integration |
| \(s = \dfrac{4}{3}t^3 - 10t^2 + 21t\, (+c)\) | A1 | Correct integration |
| \(\dfrac{49}{6} - \dfrac{27}{2}\) | M1 | Substitute their limits (\(1.5\) and \(3.5\)) into their integral |
| Distance \(= \dfrac{16}{3} = 5.33\text{ m}\) | A1 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4t^2 - 20t + 21 = (2t-3)(2t-7) = 0 \rightarrow t = \ldots$ | M1 | For setting $v = 0$ and attempting to solve $v = 0$ |
| $t = 1.5$ and $t = 3.5$ | A1 | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = 8t - 20$, $a(0) = \ldots$ | M1 | For using $a = dv/dt$ and evaluating for $t = 0$ |
| $a = -20$ | A1 | |
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $8t - 20 = 0$, $t = 2.5 \rightarrow v = \ldots$ or $v = (2t-5)^2 - 4$, $v_{\min} = \ldots$ | M1 | For setting $a = 0$, attempting to solve for $t$ and substituting to obtain $v$, or for attempting to complete the square on the expression for $v$ |
| $v_{\min} = -4 \text{ ms}^{-1}$ | A1 | |
## Question 5(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s = \int(4t^2 - 20t + 21)\, dt$ | M1 | For using $s = \int v\, dt$ and attempting integration |
| $s = \dfrac{4}{3}t^3 - 10t^2 + 21t\, (+c)$ | A1 | Correct integration |
| $\dfrac{49}{6} - \dfrac{27}{2}$ | M1 | Substitute their limits ($1.5$ and $3.5$) into their integral |
| Distance $= \dfrac{16}{3} = 5.33\text{ m}$ | A1 | |5 A particle $P$ moves in a straight line. It starts at a point $O$ on the line and at time $t$ s after leaving $O$ it has velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where $v = 4 t ^ { 2 } - 20 t + 21$.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $t$ for which $P$ is at instantaneous rest.
\item Find the initial acceleration of $P$.
\item Find the minimum velocity of $P$.
\item Find the distance travelled by $P$ during the time when its velocity is negative.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q5 [10]}}