| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a straightforward piecewise kinematics problem requiring standard differentiation to find velocity, solving v=0 for rest, evaluating velocities at t=6, and calculating distance with attention to direction changes. While it has multiple parts and requires careful bookkeeping, all techniques are routine M1 material with no novel problem-solving insight needed. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02a Kinematics language: position, displacement, velocity, acceleration3.02c Interpret kinematic graphs: gradient and area3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([v = 2t - 3]\) | M1 | For differentiation of \(s\) for \(0 \leqslant t \leqslant 6\) |
| \(t = 1.5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Velocity at arrival \(= 9 \text{ ms}^{-1}\) | B1 | \(t = 6\) used in \(v\) |
| \(v = -\dfrac{24}{t^2} - 0.5t\) | M1 | For differentiation of \(s\) for \(t \geqslant 6\) |
| Velocity when leaves \(= -3.67 \text{ ms}^{-1}\) | A1 | Allow \(v = -11/3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At \(t = 0\), \(s = 2\) or at \(t = 6\), \(s = 20\) | B1 | SOI |
| At \(t = 1.5\), \(s = -0.25\) | B1 | SOI |
| At \(t = 10\), \(s = 2.4\) | B1 | SOI |
| \([\text{Total distance} = 2 + 0.25 + 0.25 + 20 + (20 - 2.4)]\) | M1 | Evidence of distance rather than displacement involving all three sections, \((0,\ 1.5)\), \((1.5,\ 6)\) and \((6,\ 10)\) |
| So total distance travelled \(= 40.1\) m | A1 |
## Question 7:
### Part 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[v = 2t - 3]$ | M1 | For differentiation of $s$ for $0 \leqslant t \leqslant 6$ |
| $t = 1.5$ | A1 | |
### Part 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Velocity at arrival $= 9 \text{ ms}^{-1}$ | B1 | $t = 6$ used in $v$ |
| $v = -\dfrac{24}{t^2} - 0.5t$ | M1 | For differentiation of $s$ for $t \geqslant 6$ |
| Velocity when leaves $= -3.67 \text{ ms}^{-1}$ | A1 | Allow $v = -11/3$ |
### Part 7(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| At $t = 0$, $s = 2$ **or** at $t = 6$, $s = 20$ | B1 | SOI |
| At $t = 1.5$, $s = -0.25$ | B1 | SOI |
| At $t = 10$, $s = 2.4$ | B1 | SOI |
| $[\text{Total distance} = 2 + 0.25 + 0.25 + 20 + (20 - 2.4)]$ | M1 | Evidence of distance rather than displacement involving all three sections, $(0,\ 1.5)$, $(1.5,\ 6)$ and $(6,\ 10)$ |
| So total distance travelled $= 40.1$ m | A1 | |7 A particle moves in a straight line through the point $O$. The displacement of the particle from $O$ at time $t \mathrm {~s}$ is $s \mathrm {~m}$, where
$$\begin{array} { l l }
s = t ^ { 2 } - 3 t + 2 & \text { for } 0 \leqslant t \leqslant 6 , \\
s = \frac { 24 } { t } - \frac { t ^ { 2 } } { 4 } + 25 & \text { for } t \geqslant 6 .
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $t$ when the particle is instantaneously at rest during the first 6 seconds of its motion.\\
At $t = 6$, the particle hits a barrier at a point $P$ and rebounds.
\item Find the velocity with which the particle arrives at $P$ and also the velocity with which the particle leaves $P$.
\item Find the total distance travelled by the particle in the first 10 seconds of its motion.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q7 [10]}}