A particle moves in the xy plane in a circle centered on the origin.At a certain instant the velocity and acceleration of the particle are 6.0 M/s and (3.0 + 4)0 )m/s².What are the x and y coordinates of the particle at this moment? A)x = 0,y = −9.0 m B)x = 0,y = +7.2 m C)x = 0,y = +9.0 m D)x = 0,y = −7.2 m E)x = 6.0 m,y = −9.0 m

Question

Quizplus · Accepted Answer

Since the particle is moving in a circle centered at the origin, at any given moment the velocity vector is tangent to the circle and the acceleration vector points towards the center of the circle.

At the given instant, the velocity vector is 6.0 m/s in some direction, and the acceleration vector is (3.0,4.0) m/s^2. Since the acceleration vector points towards the center of the circle, we know that it has components in the negative x and y directions.

We can use this information to set up a system of equations:
6.0 cosθ = ax
6.0 sinθ = ay
ax = -3.0
ay = -4.0

Solving for θ, we find that θ = 233.1 degrees. 
Using this value of θ, we can find the x and y coordinates of the particle:
x = 6.0 cosθ = 6.0 cos(233.1) = -1.8
y = 6.0 sinθ = 6.0 sin(233.1) = -9.0

Therefore, the particle is at the coordinates (x,y) = (-1.8,-9.0). Rounded to one decimal, the answer is (0,-9.0) or choice A.

A Particle Moves in the Xy Plane in a Circle