CAIE M1 2024 November — Question 4 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeMulti-stage motion with algebraic unknowns
DifficultyModerate -0.3 This is a standard kinematics problem using velocity-time graphs with constant acceleration. It requires setting up equations for the three phases of motion and solving simultaneously, but follows a routine textbook approach with no novel insight needed. The algebra is straightforward once the setup is complete.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae
A bus travels between two stops, \(A\) and \(B\). The bus starts from rest at \(A\) and accelerates at a constant rate of \(a \text{ ms}^{-2}\) until it reaches a speed of \(16 \text{ ms}^{-1}\). It then travels at this constant speed before decelerating at a constant rate of \(0.75 \text{ ms}^{-2}\), coming to rest at \(B\). The total time for the journey is \(240\) s.
  1. Sketch the velocity-time graph for the bus's journey from \(A\) to \(B\). [1]
  2. Find an expression, in terms of \(a\), for the length of time that the bus is travelling with constant speed. [2]
  3. Given that the distance from \(A\) to \(B\) is \(3000\) m, find the value of \(a\). [3]
Question 4:
AnswerMarks Guidance
4(a)B1 Correct shape, starting at O and finishing on the t-axis.
1
AnswerMarks
4(b)16 64
t = , t =
1 2
AnswerMarks Guidance
a 3a*B1 Attempt at finding either the time for accelerating or for
decelerating – must be in terms of a.
16 64
T =240− + 
AnswerMarks Guidance
 a 3aDB1 OE – allow un-simplified.
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
4(c)1
3000= 16(T +240) [T =135]
AnswerMarks Guidance
2*M1 Use distance is area under the graph.
1  16 64
3000= 16 240+240− +
 
2   a 3a
16 64
135=240− + 
AnswerMarks Guidance
 a 3aDM1 Get an expression in terms of a ONLY using their T from part
(b) and solve for a – their T must have come from an expression
k k
of the form 240− 1 − 2 where k and k are positive
a a 1 2
constants.
OE e.g.
1 16 1 64
3000=16240− 16 − 16 .
2 a 2 3a
16
a=
AnswerMarks Guidance
45A1 Allow 0.356 or better.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
--- 4(a) ---
4(a) | B1 | Correct shape, starting at O and finishing on the t-axis.
1
--- 4(b) ---
4(b) | 16 64
t = , t =
1 2
a 3a | *B1 | Attempt at finding either the time for accelerating or for
decelerating – must be in terms of a.
16 64
T =240− + 
 a 3a | DB1 | OE – allow un-simplified.
2
Question | Answer | Marks | Guidance
--- 4(c) ---
4(c) | 1
3000= 16(T +240) [T =135]
2 | *M1 | Use distance is area under the graph.
1  16 64
3000= 16 240+240− +
 
2   a 3a
16 64
135=240− + 
 a 3a | DM1 | Get an expression in terms of a ONLY using their T from part
(b) and solve for a – their T must have come from an expression
k k
of the form 240− 1 − 2 where k and k are positive
a a 1 2
constants.
OE e.g.
1 16 1 64
3000=16240− 16 − 16 .
2 a 2 3a
16
a=
45 | A1 | Allow 0.356 or better.
3
Question | Answer | Marks | Guidance
A bus travels between two stops, $A$ and $B$. The bus starts from rest at $A$ and accelerates at a constant rate of $a \text{ ms}^{-2}$ until it reaches a speed of $16 \text{ ms}^{-1}$. It then travels at this constant speed before decelerating at a constant rate of $0.75 \text{ ms}^{-2}$, coming to rest at $B$. The total time for the journey is $240$ s.

\begin{enumerate}[label=(\alph*)]
\item Sketch the velocity-time graph for the bus's journey from $A$ to $B$. [1]

\item Find an expression, in terms of $a$, for the length of time that the bus is travelling with constant speed. [2]

\item Given that the distance from $A$ to $B$ is $3000$ m, find the value of $a$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2024 Q4 [6]}}
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