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(A) The object moves in a circular path with the center at the origin $(0,0)$.Given that at $x = -2 \, m$,the velocity is $\vec{v} = -(4 \, m/s) \hat{

Vedclass · Accepted Answer

$-(8 \, m/s^2) \hat{j}$

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(A) The object moves in a circular path with the center at the origin $(0,0)$.Given that at $x = -2 \, m$,the velocity is $\vec{v} = -(4 \, m/s) \hat{j}$.Since the velocity is tangent to the circular path,at $x = -2 \, m$ (which is on the negative $X$-axis),the velocity vector points in the negative $Y$-direction. This implies the object is moving clockwise.The radius of the circle is $R = 2 \, m$.The speed of the object is $v = 4 \, m/s$.For uniform circular motion,the acceleration is centripetal,directed towards the center.When the object is at $y = 2 \, m$ (which is on the positive $Y$-axis),the center is at $(0,0)$,so the acceleration vector must point towards the origin,i.e.,in the negative $Y$-direction.The magnitude of centripetal acceleration is $a_c = \frac{v^2}{R} = \frac{(4)^2}{2} = \frac{16}{2} = 8 \, m/s^2$.Therefore,the acceleration vector at $y = 2 \, m$ is $\vec{a} = -(8 \, m/s^2) \hat{j}$.

An object moves at a constant speed along a circular path in the horizontal $XY$ plane with the center at the origin. When the object is at $x = -2 \, m$,its velocity is $-(4 \, m/s) \hat{j}$. What is the object's acceleration when it is at $y = 2 \, m$?

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