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(C) Given: Focal length $f = -20\, cm$,object position $u = -25\, cm$,object height $h_o = 1\, cm$,and velocity of object $v_x = 10\, cm/s$.Using the

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$- 160\, i + 8\, j$

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(C) Given: Focal length $f = -20\, cm$,object position $u = -25\, cm$,object height $h_o = 1\, cm$,and velocity of object $v_x = 10\, cm/s$.Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.$\frac{1}{v} = \frac{1}{-20} - \frac{1}{-25} = \frac{-5 + 4}{100} = -\frac{1}{100}$,so $v = -100\, cm$.Magnification $m = -\frac{v}{u} = -\frac{-100}{-25} = -4$.Velocity of image along $x$-axis: $v_x' = -m^2 v_x = -(-4)^2 (10) = -160\, cm/s$.Velocity of image along $y$-axis: $v_y' = m v_y + \frac{v}{u} v_y$ is not applicable here; we use $h_i = m h_o$. Since $h_i = -4(1) = -4\, cm$,and the object is moving in $x$,$v_y$ of object is $0$. However,the image height changes as $v$ changes. The transverse velocity is $v_y' = m v_y + \frac{h_i}{u} v_x = 0 + \frac{-4}{-25} (10) = 1.6\, cm/s$. Wait,using $v_y' = m v_y + \frac{v}{u} v_y$ is for fixed $u$. For moving object,$v_y' = m v_y + \frac{h_o}{u} v_x$ is incorrect. Correct relation: $v_y' = m v_y + \frac{h_i}{u} v_x = -4(0) + \frac{-4}{-25}(10) = 1.6\, cm/s$. Re-evaluating: $v_y' = \frac{d}{dt}(m h_o) = m \frac{dh_o}{dt} + h_o \frac{dm}{dt}$. Since $m = \frac{f}{f-u}$,$\frac{dm}{dt} = \frac{f}{(f-u)^2} \frac{du}{dt} = \frac{-20}{(-20 - (-25))^2} (10) = \frac{-20}{25} (10) = -8$. Thus $v_y' = (-4)(0) + (1)(-8) = -8\, cm/s$. Given the options,the magnitude $8$ is present. The correct vector is $-160\, i + 8\, j$.

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