A particle, initially at rest, moves along the x-axis such that its acceleration at time t0 is given by a(t)=cos(t). At the time t=0, its position is x=3.

How do I find the velocity and position functions for the particle?

I tried integrating the equation but had to integrate a sin(t) + C, something I don't know how to do.

Please help! Thanks :D

Question

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 velocity and position functions for the particle?

I tried integrating the equation but had to integrate a sin(t) + C, something I don't know how to do.

Please help! Thanks :D

anonymous · Answer

you got the main part already. what you missed is that initially it is at rest, so this means 
$$v(0)=0$$

anonymous · Answer

so since 
$$v(t)=\sin(t)+C$$ and 
$$v(0)=0$$ you have 
$$v(0)=\sin(0)+C=0$$ and therefore 
$$C=0$$

anonymous · Answer

therefore 
$$v(t)=\sin(t)$$ exactly. now take the antiderivative again.  you get 
$$S(t)=-\cos(t)+C$$ and this time you know 
$$S(0)=-\cos(0)+C=3$$ so 
$$-1+C=3$$ and therefore 
$$C=4$$