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(B) Let the insect move with speed $v_0$ along the diagonal of the floor. The velocity components along the two adjacent walls are $v_x = v_0 \cos 45^

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) Let the insect move with speed $v_0$ along the diagonal of the floor. The velocity components along the two adjacent walls are $v_x = v_0 \cos 45^\circ$ and $v_y = v_0 \sin 45^\circ$.For a plane mirror,the velocity of the image parallel to the mirror is the same as the velocity of the object parallel to the mirror.Given that the speed of the image in the adjacent walls (parallel to the walls) is $20\sqrt{2} \ cm/s$. Since the insect moves at $45^\circ$ to the walls,the component of velocity parallel to one wall is $v_0 \cos 45^\circ = v_0 / \sqrt{2}$.Thus,$v_0 / \sqrt{2} = 20\sqrt{2} \implies v_0 = 40 \ cm/s$.The image formed on the roof is a reflection of the insect moving on the floor. Since the roof is parallel to the floor,the velocity of the image on the roof is equal to the velocity of the object on the floor.Therefore,the speed of the image on the roof with respect to the ground is $v_0 = 40 \ cm/s$.

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