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• Notes

$\begin{array}{l}{12-14 . \text { The automobile has a speed of } 80 \mathrm{ft} / \mathrm{s} \text { at point } A} \\ {\text { and an acceleration having a magnitude of } 10 \mathrm{ft} / \mathrm{s}^{2} \text { , acting in }} \\ {\text { the direction shown. Determine the radius of curvature of }} \\ {\text { the path at point } A \text { and the tangential component of }} \\ {\text { acceleration. }}\end{array}$

$v=80 \mathrm{ft} / \mathrm{s}$

$a=10 \mathrm{ft} / \mathrm{s}^{2}$

$a_{t}=a \cos 30=10 \cos 30 ≃ 8.66 \mathrm{ft} / \mathrm{s}^{2}$

$a_{n}=a \sin 30=10 \sin 30=5 ft / s^{2}$

$a_{n}=\frac{v^2}{\rho} \Rightarrow ∴ \rho= \frac{v^{2}}{a_{n}}=\frac{80^{2}}{5}=1280 \mathrm{ft}$

$\begin{array}{l}{12-125 . \text { The car travels around the circular track having a }} \\ {\text { radius of } r=300 \mathrm{m} \text { such that when it is at point } A \text { it has a }} \\ {\text { velocity of } 5 \mathrm{m} / \mathrm{s} \text { , which is increasing at the rate of }} \\ {\dot{v}=(0.06 t) \mathrm{m} / \mathrm{s}^{2} \text { , where } t \text { is in seconds. Determine the }} \\ {\text { magnitudes of its velocity and acceleration when it has }} \\ {\text { traveled one-third the way around the track. }}\end{array}$

$r=300 m$

$v=5 \mathrm{m} / \mathrm{s}$

$\dot {v}=(0.06 t) f t / s^{2}$

$s=\frac{1}{3}(2 \pi r)$

$a t=\dot v=0.06 t=\frac{d v}{d t}$

$∴ d v=a t d t$

$\int_{s}^{v} d v= \int_{0}^{t} 0.06 t \Rightarrow v-5=\frac{0.06 t^2}{2}\Rightarrow v=0.03 t^{2}+5$

$∵ V=\frac{d s}{d t} \Rightarrow ∴ d s=v d t \rightarrow \int_{0}^{s} d s= \int_{0}^{t}\left(0.03 t^{2}+5\right) d t$

$∴ S=0.01 t^{3}+5 t \Rightarrow ∵S=\frac{1}{3}(2 \pi*300)=628.3185 \mathrm{m}$

$∴ 0.01 t^{3}+5 t=628.3185 \Rightarrow 0.01t^{3}+5 t-628.3185=0$

$t=35.58s$

$∴ V=0.03(35.58)^{2}+5≃43 \mathrm{m} / \mathrm{s}$

$a_ n=\frac{v^{2}}{\rho}=\frac{43^{2}}{300}≃6.16 \mathrm{m} / \mathrm{s}^{2}$

$a_t=(0.06)(35.58)≃2.135 \mathrm{m} / \mathrm{s}^{2}$

$a=\sqrt{a_{n}^{2}+a_{t}^{2}}=6 \cdot 52 \mathrm{m} / \mathrm{s}^{2}$