Calculus
posted by Anonymous .
A particle, initially at rest, moves along the xaxis such that its acceleration at time t>0 is given by a(t)=cos(t). At the time t=0, its position is x=3.
How do I find the position function for the particle? I tried integrating the equation but got confused.

velocity is the integral of acceleration.
V= INT cos(t)= sinT + C
position is the integral of velocity..
position= INt (sinT+c)dt= cosT+ CT+ D
So at t=0, position is zero
position=cos0+c*O+ D so
3=1+D and D=4
C cannot be determined without more information. 
Thanks so much, I got that point but didn't that that was right. I guess I'll just leave as you explained. You've been super helpful. Thanks again!

What type of information would be needed? Does it matter that "the particle is moving along the xaxis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x"(t) is its acceleration."?

Yes, it matters, but as you can see from the equations, unless you know the initial velocity, its position cannot be determined. If it comes shooting out of a gate at t=0, the initial position and acceleration can be the same, but its position will be a lot different if it starts from rest. So, you need either v(a) for some a, or p(a) for some a other than 0.
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