(a) (16/3,8/3), reached at tA=8/3 and tB=2/3; (b) t=5/3, dmin=2.
Worked solution
(a) For A, y=x/2. For B, t=6−x, so y=4−2(6−x)=2x−8. Solving x/2=2x−8 gives the crossing point (16/3,8/3). Particle A reaches it at t=8/3 and particle B at t=2/3.
(b) At a common time, s=rB−rA=(6−3t,4−3t). Thus d2=(6−3t)2+(4−3t)2=18t2−60t+52. Differentiating gives 36t−60=0, so t=5/3. The positive quadratic coefficient establishes a minimum. At this time d2=2, so dmin=2.
Derives the paths and crossing point.
[2 marks]
Determines both crossing times.
[1 mark]
Forms the separation-squared model.
[2 marks]
Determines the minimising time.
[1 mark]
Determines the minimum separation.
[1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.