anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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dumbcow
  • dumbcow
first thought is if it has constant velocity .... then acceleration must be 0
anonymous
  • anonymous
Indeed, that was my first thought to but the key says \[a=-3\pi ^{2}i-4\pi ^{2}j\]
dumbcow
  • dumbcow
oh 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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anonymous
  • anonymous
Oh I see, thanks!
dumbcow
  • dumbcow
yw

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