Answer :
Answer:
[tex]t_1=\frac{5\pi}{17} -\frac{5\pi}{34} \approx 0.462 s\\\\t_2=\frac{10\pi}{17} -\frac{5\pi}{34} \approx1.39s[/tex]
Explanation:
No matter the coeficient of the cosine function, the function will always be zero as long as the following is true:
[tex]cos(t)=0\\\\for\\\\t=\pi n-\frac{\pi}{2} ,\hspace{7}n\in Z[/tex]
Now:
Rewrite 3.4 as:
[tex]3.4=\frac{17}{5}[/tex]
So:
[tex]\frac{17}{5} t= \pi n -\frac{\pi}{2} \\\\Hence\\\\t=\frac{5\pi n}{17} -\frac{5\pi}{34},\hspace{7}n\in Z[/tex]
Therefore the particle crosses the x-axis (x(t)=0) :
[tex]x(t)=4.5cos((3.4)t)=0,\hspace{10}When\\\\t=\frac{5\pi n}{17} -\frac{5\pi}{34},\hspace{7}n\in Z\\[/tex]
The first time is when n=1, so:
[tex]t_1=\frac{5\pi(1)}{17} -\frac{5\pi}{34}=\frac{5\pi}{17} -\frac{5\pi}{34} \approx 0.462 s[/tex]
And the second time is when n=2, so:
[tex]t_2=\frac{5\pi(2)}{17} -\frac{5\pi}{34}=\frac{10\pi}{17} -\frac{5\pi}{34} \approx1.39s[/tex]