Aparticle moves along a circular path over a horizontal xy coordinate system, at constant speed. at time t1 = 4.50 s, it is at point (5.80 m, 5.60 m) with velocity (3.50 m/s) j and acceleration in the positive x direction. at time t2 = 11.7 s, it has velocity (–3.50 m/s) i and acceleration in the positive y direction. what are the (a) x and (b) y coordinates of the center of the circular path? assume at both times that the particle is on the same orbit.

a) x=11.15m

b) y= 5.6m

Explanation:

This is the given data:

t1=4.5s                          t2=11.7s

V1=3.5 (j)m/s                 V2=-3.5 (i)m/s

a1 = a(i)                          a2 = a(j)

r1 = 5.8(i)+5.6(j) m

From this data we conclude that the orbit is as the one shown in the picture.

So, the position of the center will be rc = r1 + R(i) = (5.8+R)(i) + 5.6(j)

Calculating the distance traveled by the particle:

S = V*Δt = 3.5 * (11.7-4.5) = 25.2m

Since S = θ * R  and from the picture we know that θ = 3π/2

R = 5.35m

Now we calculate the center of the orbit as:

rc = r1 + R(i) = (5.8+5.35)(i)  +  5.6(j) = [11.15(i) + 5.6(j)] m

explanation:

newton's third law, for every action there is an equal and opposite reaction.

it is called kinetic energy