Answer:
After 1 seconds
Explanation:
The velocity of the particle is given by a periodic function, in the form
[tex]v_x=(2.2) sin (\pi t)[/tex] [m/s]
where
t is the time, in seconds
A periodic function is written in the form
[tex]y=Asin(\omega t)[/tex]
where
A is the amplitude (maximum displacement from equilibrium position)
[tex]\omega[/tex] is the angular frequency, which can be written as
[tex]\omega=\frac{2\pi}{T}[/tex]
where T is the period.
For the periodic function in this problem, we have
A = 2.2 m/s
[tex]\omega= \pi[/tex] [rad/s]
So the period is:
[tex]T=\frac{2\pi}{\pi}=2 s[/tex]
The particle reaches a turning point when the velocity becomes zero, and then it becomes negative.
At t = 0, the function has value of zero.
After 1/4 of period, which corresponds to 0.5 seconds, the function reaches its maximum:
[tex]v(0.5 s)=(2.2)\cdot sin(\pi \cdot 0.5)=2.2[/tex]
After 1/2 of period, which corresponds to 1 second, the function becomes zero again:
[tex]v(1s)=2.2\cdot sin (\pi \cdot 1)=0[/tex]
Therefore, the particle reaches the first turning point at t = 1 sec.
