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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.

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dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1first 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 \[a=3\pi ^{2}i4\pi ^{2}j\]

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.1oh 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
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