Posted by **Anonymous** on Monday, October 17, 2011 at 8:25pm.

A particle, initially at rest, moves along the x-axis 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.

- Calculus -
**bobpursley**, Monday, October 17, 2011 at 8:30pm
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.

- Calculus -
**Anonymous**, Monday, October 17, 2011 at 9:12pm
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!

- Calculus -
**Anonymous**, Monday, October 17, 2011 at 9:17pm
What type of information would be needed? Does it matter that "the particle is moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x"(t) is its acceleration."?

- Calculus -
**Steve**, Tuesday, October 18, 2011 at 12:29am
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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