QCE Vault / Specialist Maths Vectors and matrices — Question 6 QCAA 2025, Paper 1 · 1 mark
Browse all questions Vectors revision Original practice exam Report an issue Q6 · 2025 · Technology-free Simple familiar 1 mark
QUESTION 6 At time t t t , a particle travels with a velocity of v = ( 2 1 + t 2 ) i ^ − 2 t j ^ \mathbf v=\left(\dfrac{2}{1+t^2}\right)\hat{\mathbf i}-2t\hat{\mathbf j} v = ( 1 + t 2 2 ) i ^ − 2 t j ^ . Determine a general expression for the position vector, r \mathbf r r , of the particle during this motion. (A) r = 2 tan − 1 ( t ) i ^ − 2 j ^ + c \mathbf r=2\tan^{-1}(t)\hat{\mathbf i}-2\hat{\mathbf j}+\mathbf c r = 2 tan − 1 ( t ) i ^ − 2 j ^ + c (B) r = 2 tan − 1 ( t ) i ^ − t 2 j ^ + c \mathbf r=2\tan^{-1}(t)\hat{\mathbf i}-t^2\hat{\mathbf j}+\mathbf c r = 2 tan − 1 ( t ) i ^ − t 2 j ^ + c (C) r = 1 2 tan − 1 ( t ) i ^ − 2 j ^ + c \mathbf r=\frac12\tan^{-1}(t)\hat{\mathbf i}-2\hat{\mathbf j}+\mathbf c r = 2 1 tan − 1 ( t ) i ^ − 2 j ^ + c (D) r = 1 2 tan − 1 ( t ) i ^ − t 2 j ^ + c \mathbf r=\frac12\tan^{-1}(t)\hat{\mathbf i}-t^2\hat{\mathbf j}+\mathbf c r = 2 1 tan − 1 ( t ) i ^ − t 2 j ^ + c WORKED SOLUTION
Answer B 1 mark Worked solution
Integrate each component: ∫ 2 / ( 1 + t 2 ) d t = 2 tan − 1 t \int2/(1+t^2)dt=2\tan^{-1}t ∫ 2/ ( 1 + t 2 ) d t = 2 tan − 1 t and ∫ − 2 t d t = − t 2 \int-2t\,dt=-t^2 ∫ − 2 t d t = − t 2 . Hence r = 2 tan − 1 ( t ) i ^ − t 2 j ^ + c \mathbf r=2\tan^{-1}(t)\hat{\mathbf i}-t^2\hat{\mathbf j}+\mathbf c r = 2 tan − 1 ( t ) i ^ − t 2 j ^ + c . Worked explanation by QCE Vault; correct option verified against the QCAA answer key.
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