QCE Vault / Specialist Maths Vectors and matrices — Question 3 QCAA 2021, Paper 1 · 1 mark
Browse all questions Vectors revision Original practice exam Report an issue Q3 · 2021 · Technology-free Simple familiar 1 mark
QUESTION 3 An object has a velocity v ( t ) = e − 2 t i ^ + ( 1 t ) k ^ \mathbf v(t)=e^{-2t}\hat{\mathbf i}+\left(\frac1t\right)\hat{\mathbf k} v ( t ) = e − 2 t i ^ + ( t 1 ) k ^ , where t t t represents time ( t > 0 ) (t>0) ( t > 0 ) . The displacement r ( t ) \mathbf r(t) r ( t ) of the object could be (A) − 2 e − 2 t i ^ + ln ( t ) k ^ -2e^{-2t}\hat{\mathbf i}+\ln(t)\hat{\mathbf k} − 2 e − 2 t i ^ + ln ( t ) k ^ (B) − 2 e − 2 t i ^ − 1 t 2 k ^ -2e^{-2t}\hat{\mathbf i}-\frac1{t^2}\hat{\mathbf k} − 2 e − 2 t i ^ − t 2 1 k ^ (C) − 1 2 e − 2 t i ^ + ln ( t ) k ^ -\frac12e^{-2t}\hat{\mathbf i}+\ln(t)\hat{\mathbf k} − 2 1 e − 2 t i ^ + ln ( t ) k ^ (D) − 1 2 e − 2 t i ^ − 1 t 2 k ^ -\frac12e^{-2t}\hat{\mathbf i}-\frac1{t^2}\hat{\mathbf k} − 2 1 e − 2 t i ^ − t 2 1 k ^ WORKED SOLUTION
Answer C 1 mark Worked solution
Integrating componentwise gives ∫ e − 2 t d t = − 1 2 e − 2 t \int e^{-2t}dt=-\frac12e^{-2t} ∫ e − 2 t d t = − 2 1 e − 2 t and ∫ ( 1 / t ) d t = ln t \int(1/t)dt=\ln t ∫ ( 1/ t ) d t = ln t for t > 0 t>0 t > 0 . Hence option C is a possible displacement vector (up to an additive constant vector). Worked explanation by QCE Vault; correct option verified against the QCAA answer key.
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