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VICSpecialist MathematicsQuick questions

Unit 4: Calculus

Quick questions on Kinematics and rectilinear motion: VCE Specialist Mathematics Unit 4

8short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.

What is choose the form?
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Acceleration depends on xx, so use a=ddx ⁣(12v2)a = \dfrac{\mathrm{d}}{\mathrm{d}x}\!\left(\tfrac12 v^2\right):
What is apply the condition?
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At x=0x = 0, v=6v = 6, so 12(6)2=−2(0)2+c\tfrac12(6)^2 = -2(0)^2 + c, giving c=18c = 18. Thus
What is find where the particle is at rest?
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Set v=0v = 0: 36−4x2=036 - 4x^2 = 0, so x2=9x^2 = 9 and x=±3x = \pm 3.
What is check?
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At x=0x = 0, v2=36v^2 = 36 so v=6v = 6, matching. Differentiating v2=36−4x2v^2 = 36 - 4x^2 implicitly: 2vdvdx=−8x2v\frac{\mathrm{d}v}{\mathrm{d}x} = -8x, so vdvdx=−4x=av\frac{\mathrm{d}v}{\mathrm{d}x} = -4x = a, confirming the acceleration relationship. The particle oscillates between x=−3x = -3 and x=3x = 3.
What is sign of acceleration and speeding up?
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A particle slows down when vv and aa have opposite signs, even if a>0a > 0. Do not assume positive acceleration means speeding up.
What is q1?
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A particle has x=t3−6t2+9tx = t^3 - 6t^2 + 9t. Find its velocity and the times it is at rest. [3 marks]
What is q2?
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Given a=6ta = 6t with v(0)=2v(0) = 2 and x(0)=0x(0) = 0, find x(t)x(t). [3 marks]
What is q3?
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A particle has v2=16−x2v^2 = 16 - x^2. Find its acceleration in terms of xx. [2 marks]

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