To find the velocity and acceleration vectors for uniform circular motion and to recognize that this acceleration is the centripetal acceleration. suppose that a particle's position is given by the following expression: r(t)=r[cos(ωt)i^+sin(ωt)j^] =rcos(ωt)i^+rsin(ωt)j^.the particle's motion can be described by ) an ellipse starting at time t=0 on the positive x axis(b) an ellipse starting at time t=0 on the positive y axis(c) a circle starting at time t=0 on the positive x axis(d) a circle starting at time t=0 on the positive y axis

(C) a circle starting at time t=0 on the positive x axis

Explanation:

particle's position is

r(t)=R[cos(ωt)i^+sin(ωt)j^] =Rcos(ωt)i^+Rsin(ωt)j^

this is a parametric equation of a circle, because the axis at x and y are the same = R.

for t=0:

r=Ri^

so: circle starting at time t=0 on the positive x axis

On the other hand:

The value of the magnitude of the acceleration is:

we can recognise that this represent the centripetal acceleration.

magnitude and direction

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)