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(B) Given: Velocity of object $\vec{v}_o = 10 \cos(60^\circ) \hat{i} + 10 \sin(60^\circ) \hat{j} = 5 \hat{i} + 5\sqrt{3} \hat{j} 	ext{ cm/s}$.Velocit

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$ - 3i + 5\sqrt 3 \hat j$

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(B) Given: Velocity of object $\vec{v}_o = 10 \cos(60^\circ) \hat{i} + 10 \sin(60^\circ) \hat{j} = 5 \hat{i} + 5\sqrt{3} \hat{j} 	ext{ cm/s}$.Velocity of mirror $\vec{v}_m = 2 \hat{i} 	ext{ cm/s}$.The velocity of the image $\vec{v}_i$ is given by the formula: $\vec{v}_{i,x} = 2\vec{v}_{m,x} - \vec{v}_{o,x}$ and $\vec{v}_{i,y} = \vec{v}_{o,y}$.Calculating the x-component: $\vec{v}_{i,x} = 2(2) - 5 = 4 - 5 = -1 	ext{ cm/s}$.Calculating the y-component: $\vec{v}_{i,y} = 5\sqrt{3} 	ext{ cm/s}$.Thus,the velocity of the image with respect to the ground is $\vec{v}_i = -1 \hat{i} + 5\sqrt{3} \hat{j} 	ext{ cm/s}$.The velocity of the image with respect to the mirror is $\vec{v}_{im} = \vec{v}_i - \vec{v}_m = (-1 \hat{i} + 5\sqrt{3} \hat{j}) - (2 \hat{i}) = -3 \hat{i} + 5\sqrt{3} \hat{j} 	ext{ cm/s}$.

Find the velocity of the image with respect to the mirror.

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