Vector calculus — Question 9
Original QCE Vault practice · 6 marks
Q9 · Practice questionComplex unfamiliar6 marks
QUESTION 9 (6 marks)
A particle moves in a plane with acceleration
a(t)=−2i^+6tj^,t≥0.Initially, v(0)=4i^−3j^ and r(0)=i^+2j^.
a)Determine the position vector r(t).
b)Show that at t=1 the particle crosses the x-axis with horizontal velocity.
WORKED SOLUTION
Practice marking scheme
6 marksANSWERr(t)=(1+4t−t2)i^+(t3−3t+2)j^ Sample response and mark allocation
v(t)=(−2t+C1)i^+(3t2+C2)j^.
Integrates acceleration.[1 mark]
v(t)=(4−2t)i^+(3t2−3)j^.
Uses the initial velocity.[1 mark]
r(t)=(4t−t2+D1)i^+(t3−3t+D2)j^.
Integrates velocity.[1 mark]
r(t)=(1+4t−t2)i^+(t3−3t+2)j^.
Uses the initial position.[1 mark]
r(1)=4i^, so its
y-coordinate is zero.
Shows that the particle is on the x-axis.[1 mark]
v(1)=2i^, which has no vertical component.
Shows that the velocity is horizontal.[1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabusHow many marks did you earn?Compare your working with the guide above.
Related questions
- Q32 · Original practice · 7 marks
Two particles move for t≥0 with positions rA(t)=(2t,t),rB(t)=(6−t,4−2t). - Q34 · Original practice · 6 marks
A particle follows the path r(t)=(3cost,2sint),0≤t≤2π. - Q36 · Original practice · 6 marks
A projectile is launched from level ground at 20 ms−1 at 45∘ above the horizontal. Take g=10 ms−2 and ignore air resistance. - Q38 · Original practice · 5 marks
A particle has position r(t)=5(cos2t,sin2t).