Vector calculus — Question 141
Original QCE Vault practice · 6 marks
Q141 · Practice questionTechnology-activeComplex unfamiliar6 marks
QUESTION 141 (6 marks)
A particle has position r(t)=(t2,2t,3−t2) for t≥0. 
a)Find velocity and speed at t=1. [2 marks] c)Find the time when velocity is perpendicular to acceleration.
[2 marks] WORKED SOLUTION
Practice marking scheme
6 marksANSWERv⋅a=4t+4t=8t, so perpendicular when t=0. Worked solution
Part (a): v=(2t,2,−2t); at t=1speed=4+4+4=23. Part (b): a=(2,0,−2). Part (c): v⋅a=4t+4t=8t, so perpendicular when t=0. Part (a): completes the requested reasoning and obtains a correct result.
[2 marks]Part (b): completes the requested reasoning and obtains a correct result.
[2 marks]Part (c): completes the requested reasoning and obtains a correct result.
[2 marks]Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
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