OCR MEI M1 2007 January — Question 2 5 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Year2007
SessionJanuary
Marks5
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Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeFinding when particle at rest
DifficultyModerate -0.8 This is a straightforward differentiation and equation-solving problem. Students need to find v = dx/dt = 12 - 3t², set it equal to zero, and solve the quadratic to get t = ±2, then substitute back to find x = ±16. It requires only basic calculus and algebra with no conceptual challenges or multi-step reasoning.
Spec1.07a Derivative as gradient: of tangent to curve1.07m Tangents and normals: gradient and equations3.02f Non-uniform acceleration: using differentiation and integration
2 A particle moves along a straight line containing a point O . Its displacement, \(x \mathrm {~m}\), from O at time \(t\) seconds is given by $$x = 12 t - t ^ { 3 } , \text { where } - 10 \leqslant t \leqslant 10$$ Find the values of \(x\) for which the velocity of the particle is zero.
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(v = \frac{dx}{dt} = 12 - 3t^2\)M1 A1 Differentiation
Set \(v = 0\): \(12 - 3t^2 = 0\)M1
\(t^2 = 4\), so \(t = \pm 2\)A1 Both values
\(x = 12(2) - 2^3 = 16\) and \(x = 12(-2) - (-2)^3 = -16\)A1 Both values of \(x\) 
# Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = \frac{dx}{dt} = 12 - 3t^2$ | M1 A1 | Differentiation |
| Set $v = 0$: $12 - 3t^2 = 0$ | M1 | |
| $t^2 = 4$, so $t = \pm 2$ | A1 | Both values |
| $x = 12(2) - 2^3 = 16$ and $x = 12(-2) - (-2)^3 = -16$ | A1 | Both values of $x$ |
2 A particle moves along a straight line containing a point O . Its displacement, $x \mathrm {~m}$, from O at time $t$ seconds is given by

$$x = 12 t - t ^ { 3 } , \text { where } - 10 \leqslant t \leqslant 10$$

Find the values of $x$ for which the velocity of the particle is zero.

\hfill \mbox{\textit{OCR MEI M1 2007 Q2 [5]}}
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Q1 4 Q2 5 Q3 7 Q4 7 Q5 6 Q6 7 Q7 18 Q8 18