CAIE M1 2019 March — Question 6 9 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2019
SessionMarch
Marks9
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Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeVariable acceleration with initial conditions
DifficultyStandard +0.3 This is a straightforward variable acceleration problem requiring integration of a polynomial to find velocity, solving a cubic equation (which factorises nicely), and a second integration for displacement. The algebra is routine and the question guides students through the steps with clear signposting. Slightly above average due to the cubic factorisation and two-stage integration process, but all techniques are standard M1 fare.
Spec1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration
A particle moves in a straight line. It starts from rest at a fixed point \(O\) on the line. Its acceleration at time \(t\) s after leaving \(O\) is \(a\) m s\(^{-2}\), where \(a = 0.4t^3 - 4.8t^2\).
  1. Show that, in the subsequent motion, the acceleration of the particle when it comes to instantaneous rest is \(16\) m s\(^{-2}\). [6]
  2. Find the displacement of the particle from \(O\) at \(t = 5\). [3]
Question 6:
AnswerMarks
6(i) 1 
[∫0.4t3 – 4.8t2 dt]
 
AnswerMarks Guidance
 M1 Attempt to integrate a
3
AnswerMarks
v = 0.1t4 – 3.2t2 (+ c)A1
3
AnswerMarks Guidance
[v = 0 → 0.1t4 – 3.2t2 = 0]DM1 Attempt to solve v = 0, and reach the form ta/b = k
5
AnswerMarks Guidance
[t2 = 32]M1 Attempt to solve an equation of the form ta/b = k
t = 4A1
a = 16 m s–2B1
6
AnswerMarks Guidance
6(ii)3
[s =∫0.1t4 – 3.2t2 dt]M1 Attempt to integrate v
5
 5
Displacement = 0.02t5 −1.28t2
 
AnswerMarks Guidance
0A1 Correct integration.
Displacement = –9.05 m (–9.05417...)A1
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
--- 6(i) ---
6(i) |  1 
[∫0.4t3 – 4.8t2 dt]
 
  | M1 | Attempt to integrate a
3
v = 0.1t4 – 3.2t2 (+ c) | A1
3
[v = 0 → 0.1t4 – 3.2t2 = 0] | DM1 | Attempt to solve v = 0, and reach the form ta/b = k
5
[t2 = 32] | M1 | Attempt to solve an equation of the form ta/b = k
t = 4 | A1
a = 16 m s–2 | B1
6
--- 6(ii) ---
6(ii) | 3
[s =∫0.1t4 – 3.2t2 dt] | M1 | Attempt to integrate v
5
 5
Displacement = 0.02t5 −1.28t2
 
0 | A1 | Correct integration.
Displacement = –9.05 m (–9.05417...) | A1
3
Question | Answer | Marks | Guidance
A particle moves in a straight line. It starts from rest at a fixed point $O$ on the line. Its acceleration at time $t$ s after leaving $O$ is $a$ m s$^{-2}$, where $a = 0.4t^3 - 4.8t^2$.

\begin{enumerate}[label=(\roman*)]
\item Show that, in the subsequent motion, the acceleration of the particle when it comes to instantaneous rest is $16$ m s$^{-2}$. [6]

\item Find the displacement of the particle from $O$ at $t = 5$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2019 Q6 [9]}}
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