CAIE M1 2012 November — Question 2 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2012
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeDisplacement from velocity by integration
DifficultyModerate -0.8 This is a straightforward mechanics question requiring basic differentiation to find acceleration, solving a linear equation, and integration of a polynomial to find displacement. All steps are routine applications of standard techniques with no conceptual challenges or problem-solving insight required.
Spec3.02f Non-uniform acceleration: using differentiation and integration

2 A particle moves in a straight line. Its velocity \(t\) seconds after leaving a fixed point \(O\) on the line is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), where \(v = 0.2 t + 0.006 t ^ { 2 }\). For the instant when the acceleration of the particle is 2.5 times its initial acceleration,
  1. show that \(t = 25\),
  2. find the displacement of the particle from \(O\).

AnswerMarks Guidance
(i) \([a = 0.2 + 0.012t]\)M1 For differentiating to find \(a(t)\)
\([0.2 + 0.012t = 2.5 \times 0.2]\)M1 For attempting to solve \(a(t) = 2.5a(0)\)
\(t = 25\)A1 3 marks total
(ii) \([s = 0.1t^2 + 0.002t^3 \text{ (+ C)}]\)M1 For integrating to find \(s(t)\)
\([s = 0.1 \times 625 + 0.002 \times 15625]\)DM1 For using limits 0 to 25 or evaluating \(s(t)\) with \(C = 0\) (which may be implied by its absence)
Displacement is 93.75 (accept 93.7 or 93.8)A1 3 marks total
**(i)** $[a = 0.2 + 0.012t]$ | M1 | For differentiating to find $a(t)$
$[0.2 + 0.012t = 2.5 \times 0.2]$ | M1 | For attempting to solve $a(t) = 2.5a(0)$
$t = 25$ | A1 | 3 marks total | AG

**(ii)** $[s = 0.1t^2 + 0.002t^3 \text{ (+ C)}]$ | M1 | For integrating to find $s(t)$
$[s = 0.1 \times 625 + 0.002 \times 15625]$ | DM1 | For using limits 0 to 25 or evaluating $s(t)$ with $C = 0$ (which may be implied by its absence)
Displacement is 93.75 (accept 93.7 or 93.8) | A1 | 3 marks total

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2 A particle moves in a straight line. Its velocity $t$ seconds after leaving a fixed point $O$ on the line is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where $v = 0.2 t + 0.006 t ^ { 2 }$. For the instant when the acceleration of the particle is 2.5 times its initial acceleration,\\
(i) show that $t = 25$,\\
(ii) find the displacement of the particle from $O$.

\hfill \mbox{\textit{CAIE M1 2012 Q2 [6]}}