Suppose while sitting in a parked car,you notice a jogger approaching towards you in the side view mirror of radius of curvature $R = 2 \; m$. If the jogger is running at a speed of $5 \; m s^{-1}$,how fast does the image of the jogger appear to move when the jogger is $(a) \; 39 \; m$,$(b) \; 29 \; m$,$(c) \; 19 \; m$,and $(d) \; 9 \; m$ away?

(N/A) The mirror formula is given by $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,which implies $v = \frac{fu}{u-f}$.
For a convex mirror,$R = 2 \; m$,so the focal length $f = \frac{R}{2} = 1 \; m$.
Since the jogger is moving at a constant speed of $5 \; m s^{-1}$,we calculate the image position $v_1$ at distance $u$ and $v_2$ at distance $u' = u + 5$ (since the jogger moves $5 \; m$ towards the mirror in $1 \; s$). The speed of the image is $|v_1 - v_2| / 1 \; s$.
$(a)$ For $u = -39 \; m$,$v_1 = \frac{1(-39)}{-39-1} = \frac{39}{40} \; m$. For $u' = -34 \; m$,$v_2 = \frac{1(-34)}{-34-1} = \frac{34}{35} \; m$. Speed $= |\frac{39}{40} - \frac{34}{35}| = |\frac{1365 - 1360}{1400}| = \frac{5}{1400} = \frac{1}{280} \; m s^{-1}$.
$(b)$ For $u = -29 \; m$,$v_1 = \frac{29}{30} \; m$. For $u' = -24 \; m$,$v_2 = \frac{24}{25} \; m$. Speed $= |\frac{29}{30} - \frac{24}{25}| = |\frac{145 - 144}{150}| = \frac{1}{150} \; m s^{-1}$.
$(c)$ For $u = -19 \; m$,$v_1 = \frac{19}{20} \; m$. For $u' = -14 \; m$,$v_2 = \frac{14}{15} \; m$. Speed $= |\frac{19}{20} - \frac{14}{15}| = |\frac{57 - 56}{60}| = \frac{1}{60} \; m s^{-1}$.
$(d)$ For $u = -9 \; m$,$v_1 = \frac{9}{10} \; m$. For $u' = -4 \; m$,$v_2 = \frac{4}{5} \; m$. Speed $= |\frac{9}{10} - \frac{4}{5}| = |\frac{9 - 8}{10}| = \frac{1}{10} \; m s^{-1}$.