A particle is moving at a constant speed in a circular trajectory centered at the origin of an xy-coordinate system. At one point ([tex]x=4 \, \text{m}[/tex], [tex]y=0 \, \text{m}[/tex]), the particle has a velocity of [tex]-5.0 \, \text{j} \, \text{m/s}[/tex]. Determine the acceleration of the particle.

A. [tex]-6.25 \, \text{i} \, \text{m/s}^2[/tex]
B. [tex]6.25 \, \text{i} \, \text{m/s}^2[/tex]
C. [tex]6.25 \, \text{j} \, \text{m/s}^2[/tex]
D. [tex]6.25 \, (-1) \, \text{m/s}^2[/tex]

The particle is moving in a circle and hence is undergoing centripetal acceleration. The acceleration is towards the center of the circle which in this case is the positive x-direction. This yields an answer of (2) 6.25 i m/s^2.

Explanation:

The particle is moving at a constant speed on a circular path which suggests that it is under a centripetal or radial acceleration. This centripetal acceleration always points toward the center of the circle. The equation for centripetal acceleration is a = v^2/r where v is the velocity and r is the radius of the circle. In this case, velocity v=5 m/s, and the radius of the circle r is the distance from the origin to the particle, which in this case is 4 m. Substituting these values into the equation yields an acceleration of 6.25 m/s^2. However, because the velocity vector is given as negative in the y-direction, and the acceleration is always directed toward the center of the circle, the acceleration is in the positive x-direction, so the answer is (2) 6.25 i m/s^2.

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