Vectors and matrices — Question 17

QCAA 2024, Paper 2 · 6 marks

Q17 · 2024 · Technology-activeComplex familiar6 marks

QUESTION 17 (6 marks)

An object moves with a constant speed of

v

in a circular path. The position vector of the object is given by

\mathbf r=r\cos(\omega t)\hat{\mathbf i}+r\sin(\omega t)\hat{\mathbf j}

where -

r

is the radius (metres) of the circle -

\omega

is the angular velocity (radians per second) -

t

is the time (seconds) of motion for

t\ge0

Use vector calculus to prove that the magnitude of the acceleration of the object is

|\mathbf a|=\dfrac{v^2}{r}

WORKED SOLUTION

6 marks

ANSWER

|\mathbf a|=\dfrac{v^2}{r}

Worked solution

Given

\mathbf r=r\cos(\omega t)\hat{\mathbf i}+r\sin(\omega t)\hat{\mathbf j},

\mathbf v=\frac{d\mathbf r}{dt}=-r\omega\sin(\omega t)\hat{\mathbf i}+r\omega\cos(\omega t)\hat{\mathbf j},

and

\mathbf a=\frac{d\mathbf v}{dt}=-r\omega^2\cos(\omega t)\hat{\mathbf i}-r\omega^2\sin(\omega t)\hat{\mathbf j}.

The speed is

v=|\mathbf v|=\sqrt{r^2\omega^2\bigl(\sin^2(\omega t)+\cos^2(\omega t)\bigr)}=r\omega.

Also,

|\mathbf a|=\sqrt{r^2\omega^4\bigl(\cos^2(\omega t)+\sin^2(\omega t)\bigr)}=r\omega^2.

Since

v=r\omega

|\mathbf a|=r\omega^2=\frac{(r\omega)^2}{r}=\frac{v^2}{r}.

correctly determines the velocity vector

[1 mark]

determines an acceleration vector

[1 mark]

determines a simplified expression for the modulus of

\mathbf v

[1 mark]

determines an expression for the modulus of

\mathbf a

[1 mark]

correctly completes the proof based on prior evidence

[1 mark]

shows logical organisation of a fully attempted proof, communicating key steps

[1 mark]

One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.

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