Example

Question:

Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of \( R = 2~\text{m} \). If the jogger is running at a speed of \( 5~\text{m/s} \), 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?

Solution:

From the mirror equation: \[ v = \frac{fu}{u - f} \] For a convex mirror (\( R = 2~\text{m} \); \( f = 1~\text{m} \)):
For \( u = -39~\text{m} \): \[ v = \frac{(-39) \times 1}{-39 - 1} = \frac{39}{40}~\text{m} \] After 1 second (\( u = -34~\text{m} \)), image position is \( v = \frac{34}{35}~\text{m} \).
Shift in position in 1 s: \[ \frac{39}{40} - \frac{34}{35} = \frac{1365 - 1360}{1400} = \frac{5}{1400} = \frac{1}{280}~\text{m} \] Average speed of the image between 39 m and 34 m: \( \frac{1}{280}~\text{m/s} \).

Similarly, for: - \( u = -29~\text{m} \), speed \( = \frac{1}{150}~\text{m/s} \) - \( u = -19~\text{m} \), speed \( = \frac{1}{60}~\text{m/s} \) - \( u = -9~\text{m} \), speed \( = \frac{1}{10}~\text{m/s} \)
Although the jogger moves at a constant speed, the speed of the image appears to increase substantially as the jogger moves closer to the mirror. This phenomenon can also be seen for moving vehicles in rear-view mirrors.