Given the velocity v=ds/dt and the initial position of a body moving along a coordinate line, find the body's position at time t. v=sin(pi*t), s(-3)=0

Question

Given the velocity v=ds/dt and the initial position of a body moving along a coordinate line, find the body's position at time t.  v=sin(pi*t), s(-3)=0

anonymous · Answer

$$ds/dt = \sin(\pi*t)$$
$$\int\limits_{0}^{s} ds= \int\limits_{0}^{t}\sin(\pi*t)$$
$$s = (-1/\pi) * \cos(\pi*t) +C$$
$$s(-3) = 0; 0 = \cos(-3\pi) +C; C = 1$$
$$s = -(1/\pi)\cos(\pi*t) +1$$

anonymous · Answer

I don't get that at all.  I have never seen those squiggle things in my life.  The example that we were viewing to help us solve this type of problem did NOT use those pretty things, whatever they are.  Any chance you could use words.  The answer is right, btw, I just can't figure out why sometimes the vaslue between the two fractions is sometimes positive and sometimes negative.  Our answers give us the solution like this  -cos(pi*t)/pi + 1/pi.  Or -1/pi.  Any simple explanation on why the sign there is sometimes positive and sometimes negative?  i'm not really good at this, first year calc online.  it's really hard.

anonymous · Answer

Dude, the "squiggly" things are what you use to show integration.  If you don't know integration, you're going to have to study up on your material.

In the MOST BASIC DEFINITION, an integral is the anti-derivative (a derivative taken backwards)

anonymous · Answer

OMG, what am I ever gonna do?  Seriously, our textbook has not even gotten to that yet.  So I will learn by studying anti-derivatives?  I am basically teaching myself this online, and most of it is BRAND NEW.  No anti-derivatives yet!  At least not in the chapter that we are in.  it's mentioning corollaries.  Ring a bell, dude?  BTW I'm a dudette!