anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[ds/dt = \sin(\pi*t)\] \[\int\limits_{0}^{s} ds= \int\limits_{0}^{t}\sin(\pi*t)\] \[s = (-1/\pi) * \cos(\pi*t) +C\] \[s(-3) = 0; 0 = \cos(-3\pi) +C; C = 1\] \[s = -(1/\pi)\cos(\pi*t) +1\]
anonymous
  • anonymous
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
  • anonymous
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)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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!

Looking for something else?

Not the answer you are looking for? Search for more explanations.