A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
A particle moves around a circle x^2+y^2=25 at constant speed, making one revolution in 2 seconds. Find its acceleration when it is at (3,4)
I have absolutly no idea how to transform x^2+y^2=25 into a vectorequation on the from \[r= x(t) i + y(t)j+z(t)k\], it would be very much appreciated with some help.
anonymous
 3 years ago
A particle moves around a circle x^2+y^2=25 at constant speed, making one revolution in 2 seconds. Find its acceleration when it is at (3,4) I have absolutly no idea how to transform x^2+y^2=25 into a vectorequation on the from \[r= x(t) i + y(t)j+z(t)k\], it would be very much appreciated with some help.

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0first thought is if it has constant velocity .... then acceleration must be 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Indeed, that was my first thought to but the key says \[a=3\pi ^{2}i4\pi ^{2}j\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok, to define circle in vector form use sin and cos x(t) = Acos(Bt) y(t) = Asin(Bt) A is radius , B is speed (radians per sec) A = 5 period = 2 > B = 2pi/2 = pi x(t) = 5cos(pi*t) y(t) = 5sin(pi*t) now find 2nd derivatives to get accelerations
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.