CAIE M1 2024 March — Question 1 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionMarch
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeDisplacement-time graph interpretation or sketching
DifficultyEasy -1.3 This is a straightforward travel graphs question requiring basic interpretation of a displacement-time graph. Part (a) involves simple calculation using distance/time (gradient of a line segment), and part (b) requires converting gradients to a velocity-time sketch—both are routine skills with no problem-solving or conceptual challenges beyond standard M1 content.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area

\includegraphics{figure_1} The displacement of a particle at time \(t\) s after leaving a fixed point \(O\) is \(s\) m. The diagram shows a displacement-time graph which models the motion of the particle. The graph consists of 4 straight line segments. The particle travels 50 m in the first 10 s, then travels at \(2\) m s\(^{-1}\) for a period of 10 s. The particle then comes to rest for a period of 20 s, before returning to its starting point when \(t = 60\).
  1. Find the velocity of the particle during the last 20 s of its motion. [2]
  2. Sketch a velocity-time graph for the motion of the particle from \(t = 0\) to \(t = 60\). [3]

Question 1:

AnswerMarks Guidance
1(a)Total distance travelled = 50+210 =70  B1
x−50
e.g. =2 →x=2(20−10)+50.
20−10
70
Velocity =− = −3.5m s–1
AnswerMarks Guidance
20B1FT Oe.
Ft their distance 50.
Must be negative.
Do not ISW if (e.g.) velocity = 3.5.
Do allow “3.5ms–1, directed towards O.”
2

AnswerMarks Guidance
1(b)Velocity 5 ms–1 for 0 t 10 B1
B1Stepped diagram with four horizontal lines segments.
Ignore vertical line segments.
AnswerMarks
B1FTAll correct.
10, 20, 40 and 60 indicated on the t-axis.
Their 5, 2 and their –3.5 (allow –3 and –4 is
acceptable too with line segment halfway between
them) indicated on the v-axis, corresponding to the
position of the horizontal line segments.
FT their –3.5 ms–1 and/or FT their 5 ms–1.
If their answer to (a) is positive, allow use of
negative their answer for this mark.
3
AnswerMarks Guidance
QuestionAnswer Marks
Question 1:
--- 1(a) ---
1(a) | Total distance travelled = 50+210 =70  | B1 | Any explicit expression for the distance,
x−50
e.g. =2 →x=2(20−10)+50.
20−10
70
Velocity =− = −3.5m s–1
20 | B1FT | Oe.
Ft their distance 50.
Must be negative.
Do not ISW if (e.g.) velocity = 3.5.
Do allow “3.5ms–1, directed towards O.”
2
--- 1(b) ---
1(b) | Velocity 5 ms–1 for 0 t 10 | B1 | May be seen on diagram.
B1 | Stepped diagram with four horizontal lines segments.
Ignore vertical line segments.
B1FT | All correct.
10, 20, 40 and 60 indicated on the t-axis.
Their 5, 2 and their –3.5 (allow –3 and –4 is
acceptable too with line segment halfway between
them) indicated on the v-axis, corresponding to the
position of the horizontal line segments.
FT their –3.5 ms–1 and/or FT their 5 ms–1.
If their answer to (a) is positive, allow use of
negative their answer for this mark.
3
Question | Answer | Marks | Guidance
\includegraphics{figure_1}

The displacement of a particle at time $t$ s after leaving a fixed point $O$ is $s$ m. The diagram shows a displacement-time graph which models the motion of the particle. The graph consists of 4 straight line segments. The particle travels 50 m in the first 10 s, then travels at $2$ m s$^{-1}$ for a period of 10 s. The particle then comes to rest for a period of 20 s, before returning to its starting point when $t = 60$.

\begin{enumerate}[label=(\alph*)]
\item Find the velocity of the particle during the last 20 s of its motion. [2]
\item Sketch a velocity-time graph for the motion of the particle from $t = 0$ to $t = 60$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2024 Q1 [5]}}