Question 6
The position vector of a particle at time $t$ is $\boldsymbol{r}(t)=5\cos(2t)\,\hat{\boldsymbol{\imath}}+5\sin(2t)\,\hat{\boldsymbol{\jmath}}$.
The magnitude of the acceleration of the particle is
Not quite. You gave the speed, $r\omega=5\times2=10$.
That’s it. Step through the hints below to see why.
Not quite. You took an extra factor of $2$ from the chain rule on the second differentiation, $5\times2^3=40$.
Not quite. You squared the radius as well, $r^2\omega^2=25\times4=100$ (the same value as $\left\lvert\boldsymbol{v}\right\rvert^2$ with no square root).
The arrows are drawn shorter so they fit. The orange arrow is drawn so it ends at the centre.
Use $\left\lvert\boldsymbol{a}\right\rvert=r\omega^2$ for uniform circular motion.
$$\left\lvert\boldsymbol{a}\right\rvert=5\times2^2=20$$
$\boldsymbol{r}(t)$ traces a circle of radius $5$ with angular speed $2$.
$$\left\lvert\boldsymbol{a}\right\rvert=r\omega^2=5\times2^2=20$$
The magnitude of the acceleration is $20$, option (B).
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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