Velocity=0 at t for Particle on x-Axis

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At time t is greater than or equal to 0, the acceleration of a particle moving on the x-axis is a(t)= t+ sin t. At t=0, the velocity of the particle is -2. For what value t will the velocity of the particle be zero?
 
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Here's a link to the Calculus Homework section.
https://www.physicsforums.com/forumdisplay.php?f=156
The sticky at the top will show you what is required for posting Homework questions.

I'll go ahead and tell you that all you basically need to know is how to find velocity (v(t)), when given acceleration (a(t)). It's quite simple and after that the question basically answers itself.
 
yeah i actually figured it out already
i couldn't figure out how to delete it
 
A mentor will probably delete or move this. I'd take note of the Homework sections and proper use though.
 
yeah i didnt know about that
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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