QCAA Specialist Mathematics Alternative Sequence Algebra of vectors in two dimensions

1 sample question with marking guides and sample answers

Q9
2021
QCAA
Paper 1
1 mark
Q9
1 mark

Car A moves with a velocity of vA=2i^5j^ m s1v_A=2\hat{\mathbf{i}}-5\hat{\mathbf{j}}\ \mathrm{m\ s^{-1}}

Car B moves with a velocity of vB=3i^+6j^ m s1v_B=3\hat{\mathbf{i}}+6\hat{\mathbf{j}}\ \mathrm{m\ s^{-1}}

The velocity of Car A relative to Car B is

A

i^11j^ m s1-\hat{\mathbf{i}}-11\hat{\mathbf{j}}\ \mathrm{m\ s^{-1}}

B

i^+j^ m s1-\hat{\mathbf{i}}+\hat{\mathbf{j}}\ \mathrm{m\ s^{-1}}

C

i^+11j^ m s1\hat{\mathbf{i}}+11\hat{\mathbf{j}}\ \mathrm{m\ s^{-1}}

D

5i^+j^ m s15\hat{\mathbf{i}}+\hat{\mathbf{j}}\ \mathrm{m\ s^{-1}}

Reveal Answer
A

i^11j^ m s1-\hat{\mathbf{i}}-11\hat{\mathbf{j}}\ \mathrm{m\ s^{-1}}

Correct Answer

The relative velocity of Car A with respect to Car B is found by subtracting B's velocity from A's: vA/B=vAvB=(23)i^+(56)j^=i^11j^ m s1v_{A/B} = v_A - v_B = (2-3)\hat{\mathbf{i}} + (-5-6)\hat{\mathbf{j}} = -\hat{\mathbf{i}} - 11\hat{\mathbf{j}}\ \mathrm{m\ s^{-1}}.

B

i^+j^ m s1-\hat{\mathbf{i}}+\hat{\mathbf{j}}\ \mathrm{m\ s^{-1}}

This incorrect answer results from a sign error when subtracting the j^\hat{\mathbf{j}} components, calculating 5(6)-5 - (-6) instead of 56-5 - 6.

C

i^+11j^ m s1\hat{\mathbf{i}}+11\hat{\mathbf{j}}\ \mathrm{m\ s^{-1}}

This is the velocity of Car B relative to Car A (vBvAv_B - v_A), which is the exact opposite of the requested relative velocity.

D

5i^+j^ m s15\hat{\mathbf{i}}+\hat{\mathbf{j}}\ \mathrm{m\ s^{-1}}

This is the sum of the two velocities (vA+vBv_A + v_B), rather than the difference required to find relative velocity.

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