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Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of R = 2 m. If the jogger is running at a speed of 5 m s^(-1), how fast 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. |
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Answer» Solution :From the mirror equation, Eq. (9.7), we get `v=(fu)/(u-F)` For convex mirror, since R = 2 m, f = 1 m. Then for `u=-39m, v=((-39)xx1)/(-39-1)=(39)/(40)m` Since the jogger moves at a constant speed of `5 m s^(-1)`, after 1 s the position of the image v `("for "u = 39 + 5 = 34)` is `(34//35 )m`. The shift in the position of image in 1 s is `(39)/(40)-(34)/(35)=(1365-1360)/(1400)=(5)/(1400)=(1)/(280)m` Therefore, the average speed of the image when the jogger is between 39 m and 34 m from the mirror, is `(1//280) m s^(-1)` SIMILARLY, it can be seen that for `u = -29 m, -19 m and -9 m`, the speed with which the image appears to move is `(1)/(150)ms^(-1),(1)/(60)ms^(-1) and (1)/(10)ms^(-1)`, respectively. Although the jogger has been moving with a constant speed, the speed of his/her image appears to INCREASE substantially as he/she moves closer to the mirror. This phenomenon can be noticed by any person sitting in a stationary car or a BUS. In case of moving vehicles, a similar phenomenon could be observed if the vehicle in the rear is moving closer with a constant speed. |
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