CAIE M1 2023 June — Question 4 8 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2023
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeMulti-stage motion with velocity-time graph given
DifficultyModerate -0.3 This is a standard velocity-time graph question requiring area calculations (distance = area under graph) and basic kinematic reasoning. Part (a) is straightforward trapezium area, part (b) uses the constraint that displacement returns to zero, and part (c) extends this with slightly more algebra. All techniques are routine for M1 students with no novel problem-solving required, making it slightly easier than average.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area
\includegraphics{figure_4} The velocity of a particle at time \(t\) s after leaving a fixed point \(O\) is \(v\) m s\(^{-1}\). The diagram shows a velocity-time graph which models the motion of the particle. The graph consists of \(5\) straight line segments. The particle accelerates to a speed of \(0.9\) m s\(^{-1}\) in a period of \(3\) s, then travels at constant speed for \(6\) s, then comes instantaneously to rest \(1\) s later. The particle then moves back and returns to rest at \(O\) at time \(T\) s.
  1. Find the distance travelled by the particle in the first \(10\) s of its motion. [2]
  2. Given that \(T = 12\), find the minimum velocity of the particle. [2]
  3. Given instead that the greatest speed of the particle is \(3\) m s\(^{-1}\), find the value of \(T\) and hence find the average speed of the particle for the whole of the motion. [4]
Question 4:
AnswerMarks
4Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
AnswerMarks
4(a)1
Distance  (610)0.9
AnswerMarks Guidance
2M1 Completely correct method for finding the total area
underneath the velocity-time graph from t = 0 to 10
only. Can be done as two triangles and a rectangle e.g.,
1 1
30.9(93)0.9 10.9 (allow a slip in
2 2
one value); need not see all three components added
together.
AnswerMarks
Distance = 7.2(m)A1
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
4(b)1
(1210)v 7.2
AnswerMarks Guidance
2 minM1 1
Setting (1210)v equal totheir (a).
2 min
AnswerMarks Guidance
Minimum velocity = 7.2 (ms–1)A1 Must be negative – allow those who solve
1
(1210)v 7.2and obtain v 7.2and then
2 min min
change to 7.2without justification.
SC B1 for assuming the triangle is isosceles.
2
AnswerMarks
4(c)1 1
(T 10)37.2 or t37.2
AnswerMarks Guidance
2 2*M1 Correct method for finding T or t. Condone sign errors
but must equate to their answer to (a).
AnswerMarks Guidance
T = 14.8A1 OE (e.g. from t + 10 = 4.8 + 10 = 14.8).
14.4
AnswerMarks Guidance
14.8DM1 2(their(a)) 2(their(a))
M1 for or .
their T 10their T
36
Average speed = (ms–1 )
AnswerMarks Guidance
37A1 OE 0.973 [For reference: 0.97297…].
SC *B1 (for T = 14.8). DM1A1 for assuming the
triangle is isosceles.
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
--- 4(a) ---
4(a) | 1
Distance  (610)0.9
2 | M1 | Completely correct method for finding the total area
underneath the velocity-time graph from t = 0 to 10
only. Can be done as two triangles and a rectangle e.g.,
1 1
30.9(93)0.9 10.9 (allow a slip in
2 2
one value); need not see all three components added
together.
Distance = 7.2(m) | A1
2
Question | Answer | Marks | Guidance
--- 4(b) ---
4(b) | 1
(1210)v 7.2
2 min | M1 | 1
Setting (1210)v equal totheir (a).
2 min
Minimum velocity = 7.2 (ms–1) | A1 | Must be negative – allow those who solve
1
(1210)v 7.2and obtain v 7.2and then
2 min min
change to 7.2without justification.
SC B1 for assuming the triangle is isosceles.
2
--- 4(c) ---
4(c) | 1 1
(T 10)37.2 or t37.2
2 2 | *M1 | Correct method for finding T or t. Condone sign errors
but must equate to their answer to (a).
T = 14.8 | A1 | OE (e.g. from t + 10 = 4.8 + 10 = 14.8).
14.4
14.8 | DM1 | 2(their(a)) 2(their(a))
M1 for or .
their T 10their T
36
Average speed = (ms–1 )
37 | A1 | OE 0.973 [For reference: 0.97297…].
SC *B1 (for T = 14.8). DM1A1 for assuming the
triangle is isosceles.
4
Question | Answer | Marks | Guidance
\includegraphics{figure_4}

The velocity of a particle at time $t$ s after leaving a fixed point $O$ is $v$ m s$^{-1}$. The diagram shows a velocity-time graph which models the motion of the particle. The graph consists of $5$ straight line segments. The particle accelerates to a speed of $0.9$ m s$^{-1}$ in a period of $3$ s, then travels at constant speed for $6$ s, then comes instantaneously to rest $1$ s later. The particle then moves back and returns to rest at $O$ at time $T$ s.

\begin{enumerate}[label=(\alph*)]
\item Find the distance travelled by the particle in the first $10$ s of its motion. [2]
\item Given that $T = 12$, find the minimum velocity of the particle. [2]
\item Given instead that the greatest speed of the particle is $3$ m s$^{-1}$, find the value of $T$ and hence find the average speed of the particle for the whole of the motion. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2023 Q4 [8]}}
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