$\displaystyle
x = 2 cos(3t+\frac{\pi}{6}\))
$
When does the particle first come to rest after t=0
Thanks Nath
At rest, velocity will be zero.Originally Posted by nath_quam
$\displaystyle velocity=\frac{dx}{dt}$
$\displaystyle \frac{dx}{dt}=-6sin(3t + \frac{\pi}{6})$
Since,$\displaystyle sin2\pi =0$
$\displaystyle 3t + \frac{\pi}{6}=2\pi$
$\displaystyle t=\frac{11\pi}{18}$
Keep Smiling
Malay
Malay's solution of 11pi/18 is approximately 1.92 so I'd say you made an error somewhere. Malay's method looks fine to me.Originally Posted by nath_quam
Edit: Now that I look at it, I wonder if Malay should've set it equal to pi since that's the next time it equals 0.
Dave
You're correct Nath, it first comes to rest(assuming it's moving along the positive x-axis) at $\displaystyle t=\frac{5{\pi}}{18}$
Left of the y-axis, it first comes to rest at $\displaystyle t=\frac{-\pi}{18}$
Add or subtract multiples of $\displaystyle \frac{\pi}{3}$ to arrive at any subsequent extrema. The corresponding y values are -2 or 2.