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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 = 2m. If the jogger is running at a speed of 5 ms^(-1) , how fast the image of the jogger appears to move when the jogger is (a) 39 m (b) 29 m (c) 19m (d) 9 m aways. |
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Answer» Solution :From the mirror equation, we get v = `(fu)/(u - f)` For cunvec mirror, SINCE R = 2m, f = 1m. Then for u = -39 m , v = `((-39)xx 1)/( - 39 - 1) = (30)/(40) `m since the jogger moves at a constant speed of 5 `ms^(-1)`, after 1s the position of the IMAGE v (for u= 39 + 5= - 34 ) is `((34)/(35))` m. The shift in the position of image in 1s 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 39m and 34 m from the mirror, is `((1)/(280) ) ms^(-1)` similarly, it can be seen that the for you u =- 29m, - 19m and -9m, the speed with which the image apprears to MOVE is `(1)/(150) ms^(-1), (1)/(60) ms^(-1) and (1)/(10) ms^(-1)` , respectively. The analysis shown that the speed of image is not same as that of object, but increases for closer positions. |
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