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Vector calculus — Question 149

Original QCE Vault practice · 6 marks

Q149 · Practice questionTechnology-freeComplex familiar6 marks

QUESTION 149 (6 marks)

Two particles move in the plane for 0≤t≤40\leq t\leq4, where tt is measured in seconds. Their position vectors, in metres, are
rA(t)=ti+(t−2)2j,rB(t)=(t+1)i+j.\mathbf r_A(t)=t\mathbf i+(t-2)^2\mathbf j,\qquad \mathbf r_B(t)=(t+1)\mathbf i+\mathbf j.
Their paths are shown. The points A0A_0 and B0B_0 indicate their positions at t=0t=0.
Equal-scale paths: A follows y=(x-2)^2 for 0<=x<=4; B follows y=1 for 1<=x<=5. Start positions A0=(0,4), B0=(1,1).
a)
Determine the two points where the paths cross. Explain why the particles never meet during the given time interval.
[3 marks]
b)
Determine the minimum distance between the particles and all times at which this minimum occurs.
[3 marks]
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