CAIE M1 2018 June — Question 7 13 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2018
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypePiecewise motion functions
DifficultyModerate -0.3 This is a straightforward multi-part kinematics question requiring standard techniques: differentiation to find maximum velocity, checking acceleration continuity, sketching a velocity-time graph, and integration to find distance. While it has multiple parts (13 marks total), each part uses routine A-level methods with no novel problem-solving required, making it slightly easier than average.
Spec1.07a Derivative as gradient: of tangent to curve1.08d Evaluate definite integrals: between limits3.02a Kinematics language: position, displacement, velocity, acceleration3.02b Kinematic graphs: displacement-time and velocity-time
A particle \(P\) moves in a straight line starting from a point \(O\). The velocity \(v\text{ m s}^{-1}\) of \(P\) at time \(t\text{ s}\) is given by $$v = 12t - 4t^2 \quad \text{for } 0 \leqslant t \leqslant 2,$$ $$v = 16 - 4t \quad \text{for } 2 \leqslant t \leqslant 4.$$
  1. Find the maximum velocity of \(P\) during the first \(2\text{ s}\). [3]
  2. Determine, with justification, whether there is any instantaneous change in the acceleration of \(P\) when \(t = 2\). [2]
  3. Sketch the velocity-time graph for \(0 \leqslant t \leqslant 4\). [3]
  4. Find the distance travelled by \(P\) in the interval \(0 \leqslant t \leqslant 4\). [5]
Question 7:
AnswerMarks
7(i)dv
[ =12−8t] or e.g. [–4[(t – 1.5)2 – 2.25]]
AnswerMarks Guidance
dtM1 For attempted differentiation of 12t−4t2 (or for alternative e.g.
completing the square)
AnswerMarks Guidance
[Maximum v when t =1.5⇒v=12×1.5−4×1.52]M1 For finding and using t
Maximum velocity is 9 (m s–1)A1
Total:3
AnswerMarks
7(ii)dv
[ =12−8t= –4]
AnswerMarks Guidance
dtM1 Finding acceleration for 0 ⩽ t ⩽ 2 when t = 2
Acceleration for 2 ⩽ t ⩽ 4 is –4
AnswerMarks Guidance
No instantaneous changeA1 Both values correct, with correct statement
Total:2
QuestionAnswer Marks
AnswerMarks Guidance
7(iii)B1 Quadratic shape (with max) for 0 ⩽ t ⩽ 2
B1Line with negative gradient from (2, …) to (4,0)
B1All correct, smooth join and key values indicated
Total:3
AnswerMarks Guidance
7(iv)Area of triangle is 8 B1
[∫(12t−4t2)dt =6t2−4t3]
AnswerMarks Guidance
3M1 Integration attempt for 0 ⩽ t ⩽ 2
[6×22 − 4×23 −6×02 + 4×03]
AnswerMarks Guidance
3 3DM1 Use of limits 0 and 2; condone absence of zero terms
40
Area under curve is or 13.3
AnswerMarks
3A1
64
Distance travelled is (m) or 21.3 (m)
AnswerMarks
3A1
Total:5 
Question 7:
--- 7(i) ---
7(i) | dv
[ =12−8t] or e.g. [–4[(t – 1.5)2 – 2.25]]
dt | M1 | For attempted differentiation of 12t−4t2 (or for alternative e.g.
completing the square)
[Maximum v when t =1.5⇒v=12×1.5−4×1.52] | M1 | For finding and using t
Maximum velocity is 9 (m s–1) | A1
Total: | 3
--- 7(ii) ---
7(ii) | dv
[ =12−8t= –4]
dt | M1 | Finding acceleration for 0 ⩽ t ⩽ 2 when t = 2
Acceleration for 2 ⩽ t ⩽ 4 is –4
No instantaneous change | A1 | Both values correct, with correct statement
Total: | 2
Question | Answer | Marks | Guidance
--- 7(iii) ---
7(iii) | B1 | Quadratic shape (with max) for 0 ⩽ t ⩽ 2
B1 | Line with negative gradient from (2, …) to (4,0)
B1 | All correct, smooth join and key values indicated
Total: | 3
--- 7(iv) ---
7(iv) | Area of triangle is 8 | B1 | (May be obtained by integrating 16 – 4t or use of uvast)
[∫(12t−4t2)dt =6t2−4t3]
3 | M1 | Integration attempt for 0 ⩽ t ⩽ 2
[6×22 − 4×23 −6×02 + 4×03]
3 3 | DM1 | Use of limits 0 and 2; condone absence of zero terms
40
Area under curve is or 13.3
3 | A1
64
Distance travelled is (m) or 21.3 (m)
3 | A1
Total: | 5
A particle $P$ moves in a straight line starting from a point $O$. The velocity $v\text{ m s}^{-1}$ of $P$ at time $t\text{ s}$ is given by
$$v = 12t - 4t^2 \quad \text{for } 0 \leqslant t \leqslant 2,$$
$$v = 16 - 4t \quad \text{for } 2 \leqslant t \leqslant 4.$$

\begin{enumerate}[label=(\roman*)]
\item Find the maximum velocity of $P$ during the first $2\text{ s}$. [3]

\item Determine, with justification, whether there is any instantaneous change in the acceleration of $P$ when $t = 2$. [2]

\item Sketch the velocity-time graph for $0 \leqslant t \leqslant 4$. [3]

\item Find the distance travelled by $P$ in the interval $0 \leqslant t \leqslant 4$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2018 Q7 [13]}}
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