CAIE M1 2018 November — Question 3 7 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2018
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeMulti-stage motion with velocity-time graph given
DifficultyModerate -0.8 This is a straightforward velocity-time graph question requiring basic kinematics: calculating acceleration from gradient, finding velocity from given conditions, and using area under graph for displacement. All techniques are standard GCSE/AS-level mechanics with clear step-by-step structure and no novel problem-solving required.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae

\includegraphics{figure_3} The velocity of a particle moving in a straight line is \(v\) m s\(^{-1}\) at time \(t\) seconds. The diagram shows a velocity-time graph which models the motion of the particle from \(t = 0\) to \(t = T\). The graph consists of four straight line segments. The particle reaches its maximum velocity \(V\) m s\(^{-1}\) at \(t = 10\).
  1. Find the acceleration of the particle during the first \(2\) seconds. [1]
  2. Find the value of \(V\). [2]
At \(t = 6\), the particle is instantaneously at rest at the point \(A\). At \(t = T\), the particle comes to rest at the point \(B\). At \(t = 0\) the particle starts from rest at a point one third of the way from \(A\) to \(B\).
  1. Find the distance \(AB\) and hence find the value of \(T\). [4]

Question 3:

AnswerMarks Guidance
3(i)Acceleration = –1 m s–2 B1
1

AnswerMarks Guidance
3(ii)[V/4 = 1 or (V + 2)/6 = 1] M1
to find V
AnswerMarks
V = 4A1
2

AnswerMarks Guidance
3(iii)[Distance = Area = ½ (6 + 2) × 2 = 8] M1
Distance AB = 3 × 8 = 24 mA1
[½ × (T – 6) × 4 = 24]M1 Attempt to find the distance travelled from t = 6 to t = T and set up an
equation for T
AnswerMarks
T = 18A1
4
AnswerMarks Guidance
QuestionAnswer Marks
Question 3:
--- 3(i) ---
3(i) | Acceleration = –1 m s–2 | B1 | Allow deceleration = 1 m s–2
1
--- 3(ii) ---
3(ii) | [V/4 = 1 or (V + 2)/6 = 1] | M1 | Use of gradient of line between t = 4 and t = 10 or use of similar triangles
to find V
V = 4 | A1
2
--- 3(iii) ---
3(iii) | [Distance = Area = ½ (6 + 2) × 2 = 8] | M1 | Attempt distance travelled in first 6 seconds
Distance AB = 3 × 8 = 24 m | A1
[½ × (T – 6) × 4 = 24] | M1 | Attempt to find the distance travelled from t = 6 to t = T and set up an
equation for T
T = 18 | A1
4
Question | Answer | Marks | Guidance
\includegraphics{figure_3}

The velocity of a particle moving in a straight line is $v$ m s$^{-1}$ at time $t$ seconds. The diagram shows a velocity-time graph which models the motion of the particle from $t = 0$ to $t = T$. The graph consists of four straight line segments. The particle reaches its maximum velocity $V$ m s$^{-1}$ at $t = 10$.

\begin{enumerate}[label=(\roman*)]
\item Find the acceleration of the particle during the first $2$ seconds. [1]

\item Find the value of $V$. [2]
\end{enumerate}

At $t = 6$, the particle is instantaneously at rest at the point $A$. At $t = T$, the particle comes to rest at the point $B$. At $t = 0$ the particle starts from rest at a point one third of the way from $A$ to $B$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the distance $AB$ and hence find the value of $T$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2018 Q3 [7]}}