Question

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.

Answer 1

first thought is if it has constant velocity .... then acceleration must be 0

Answer 2

Indeed, that was my first thought to but the key says \[a=-3\pi ^{2}i-4\pi ^{2}j\]

Answer 3

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

Answer 4

Oh I see, thanks!

Answer 5

yw