Answer
$9.51\ m$
Work Step by Step
$\vec{a}=a_{x}\hat{\mathrm{i}}+a_{y}\hat{\mathrm{j}} \qquad $(unit vector notation of vector $\vec{a}$)
The components are given by
$ a_{x}=a\cos\theta$ and $ a_{y}=a\sin\theta,\ \quad$ (3-5)
where $\theta$ is the angle between the positive direction of the $x$ axis and the direction of $\vec{a}$.
$a=\sqrt{a_{x}^{2}+a_{y}^{2}}$ and $\displaystyle \tan\theta=\frac{a_{y}}{a_{x}} \qquad$ (3-6)
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Given the magnitude and angle, using (3-5),
$ a_{x}=a\cos\theta =(17.0m)(\cos 18.0^{o})=9.51$ m
