Question

# The position of a dragonfly that is flying parallel to the ground is given as a function of time by $\vec{r}= \left[2.90 \mathrm{m}+\left(0.0900 \mathrm{m} / \mathrm{s}^{2}\right) t^{2} \hat{\imath}-\left(0.0150 \mathrm{m} / \mathrm{s}^{3}\right) t^{3}\right]$. At what value of t does the velocity vector of the dragonfly make an angle of $30.0^{\circ}$ clockwise from the +x-axis?

Solution

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(a) First we derive an expression for the velocity by differentiating the position vector with respect to time. Second we use the components of the velocity vector with the tangent function to get $t$:

$\vec{v}=\dfrac{d\vec{r}}{dt}=\dfrac{d[(2.9+0.09t^2)\hat{i}-(0.015t^3)\hat{j}]}{dt}=(0.18t)\hat{i}-(0.045t^2)\hat{j}$

$\tan\theta=\dfrac{v_y}{v_x}$

Now the desired angle the dragonfly makes with the $+x$ axis is $30^{\circ}$ clockwise, so $\theta=-30^{\circ}$

$\tan({-30}^{\circ})=\dfrac{-.045t^2}{0.18t}$

$t=\dfrac{0.18\tan({-30}^{\circ})}{-0.045}=02.31\,s$

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