| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Multi-stage motion with velocity-time graph given |
| Difficulty | Moderate -0.3 This is a straightforward velocity-time graph question requiring basic SUVAT understanding and area calculation. Part (a) involves setting two accelerations equal (simple algebra), and part (b) requires finding areas under the graph (trapezoids/triangles). While multi-stage, it's a standard M1 exercise with no conceptual challenges beyond routine application of v-t graph principles. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{20-6}{50-T} = \frac{20}{5}\) or \(20 = 6 + \frac{20}{5(50-T)}\) | M1 | Equate the accelerations and set up an equation in \(T\). Allow correct use of *their* incorrect \(\frac{20}{5}\). |
| \(T = 46.5\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distance \(= \frac{1}{2} \times 5 \times 20 + 20 \times 20 + \frac{1}{2} \times 5 \times (20+6)\) \(+ 6 \times (T-30) + \frac{1}{2} \times (50-T) \times (20+6) + \frac{1}{2} \times 10 \times 20\) \([= 50 + 400 + 65 + 99 + 45.5 + 100]\) OR Distance \(= \frac{1}{2} \times 20 \times (60+45) - \frac{1}{2} \times 14 \times (25 + T - 30)\) \([= 1050 - 290.5]\) | M1 | Attempt to find total distance using areas. Allow with \(T\) not yet substituted. Allow one error in use of area formulae or omission of only one of the areas: \(0\)–\(5\), \(5\)–\(25\), \(25\)–\(30\), \(30\)–\(T\), \(T\)–\(50\), \(50\)–\(60\). |
| Total distance travelled \(= 759.5\) m | A1 FT | FT *their* \(T\) value: Provided \(30 < T < 50\) and distance \(= 1085 - 7T\) |
| 2 |
## Question 1:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{20-6}{50-T} = \frac{20}{5}$ or $20 = 6 + \frac{20}{5(50-T)}$ | M1 | Equate the accelerations and set up an equation in $T$. Allow correct use of *their* incorrect $\frac{20}{5}$. |
| $T = 46.5$ | A1 | |
| | **2** | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distance $= \frac{1}{2} \times 5 \times 20 + 20 \times 20 + \frac{1}{2} \times 5 \times (20+6)$ $+ 6 \times (T-30) + \frac{1}{2} \times (50-T) \times (20+6) + \frac{1}{2} \times 10 \times 20$ $[= 50 + 400 + 65 + 99 + 45.5 + 100]$ OR Distance $= \frac{1}{2} \times 20 \times (60+45) - \frac{1}{2} \times 14 \times (25 + T - 30)$ $[= 1050 - 290.5]$ | M1 | Attempt to find total distance using areas. Allow with $T$ not yet substituted. Allow one error in use of area formulae or omission of only one of the areas: $0$–$5$, $5$–$25$, $25$–$30$, $30$–$T$, $T$–$50$, $50$–$60$. |
| Total distance travelled $= 759.5$ m | A1 FT | FT *their* $T$ value: Provided $30 < T < 50$ and distance $= 1085 - 7T$ |
| | **2** | |1\\
\includegraphics[max width=\textwidth, alt={}, center]{083d3e44-1e42-461f-aa8d-a1a22047a47e-02_611_1351_260_397}
The diagram shows a velocity-time graph which models the motion of a car. The graph consists of six straight line segments. The car accelerates from rest to a speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ over a period of 5 s , and then travels at this speed for a further 20 s . The car then decelerates to a speed of $6 \mathrm {~ms} ^ { - 1 }$ over a period of 5 s . This speed is maintained for a further $( T - 30 ) \mathrm { s }$. The car then accelerates again to a speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ over a period of $( 50 - T ) \mathrm { s }$, before decelerating to rest over a period of 10 s .
\begin{enumerate}[label=(\alph*)]
\item Given that during the two stages of the motion when the car is accelerating, the accelerations are equal, find the value of $T$.
\item Find the total distance travelled by the car during the motion.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q1 [4]}}