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**DarkestEvil** My class is doing a test review for homework, and I was doing well so far until I bumped into this:

A particle moves along the x-axis in such a way that its acceleration at time t for t (greater than or equal to) 0 is given by a(t)=2 cos(t). At time t=0, the velocity of the particle is v(0)= -1 and its position is x(0)= 0.

a. Write an equation for the velocity for the velocity v(t) of the particle.

Already found this, it was: v(t)=2 sin t -1

b. Write an equation for the position x(t) of the particle.

Found this also, which was: x(t)= -2 cos t - t +2

c. For what values of t, 0< t < $\displaystyle \pi$, is the particle at rest? (less thans are less than or equal to).

The problem I have with this is isolating t using: -1= -2 cos t - t +2

to find when the particle is at rest, set v(t) = 0, not x(t).

I already did a similar problem, but it was much simpler to isolate t to one side and solve for it. (2$\displaystyle \pi $-2$\displaystyle \pi $sin 2$\displaystyle \pi$t = 0 ----> t=0.25)

Another little question while I'm on the subject, how do I $\displaystyle \frac{d}{dx} tan (4x + 9x^2)? $ I'm too used to single terms.

chain rule ... $\displaystyle \textcolor{red}{\frac{d}{dx}\tan{u} = \sec^2{u} \cdot \frac{du}{dx}}$