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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 ms^(-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 :`rArr` Side mirror of car is convex and as per mirror equaton, `1/F = 1/U + 1/v` `therefore 1/v = 1/f - 1/u = (u-f)/(fu)` `therefore v = (fu)/(u-f)` and radius of curvature R = 2 cm hence focal lenght f =1 m Constant SPEED of person `v = 5 ms^(-1)` (toward car) Distance covered by person toward mirror `=vt` `= 5 xx 1` `= 5m` (a) Image distance for `u _1 = -39m, ` `v_1 = (fu_1)/(u_1 - f) = ((1)(-39))/((-39)-1) = 39/40 m ` distance of person from mmirror after 1 s, `u_2 =- 39 + 5 =- 34 m` `therefore` Image distance after 1s, `u_2 = - 39+ 5 = - 34m ` `therefore` Image distance after 1 s, `therefore v_2 = (fu_2)/(u_2-f) = ((1)(-34))/(-34 - 1) = (34)/(35) m` `thereforev_2 = (fu_2)/(u_2 - f) = ((1)(-34))/(-34-1) = (34)/(35) m` `therefore` Speed of image after 1 s, ` = (v_1 -v_2)/(t)` `= (39/40 - 34/35)/(1) = (1365 - 1360)/(1400)` `= (5)/(1400) = (1)/(280) ms^(-1)` (b) For `mu_1 = -29 m` `v_1 = (fu_1)/(u_1 -f) = ((1)(-29))/(-29-1) = (29)/(30) m ` distance of person from car after 1 s, `u_2 = -29 + 5 = -24 m` `therefore` Image distance after 1 s, `v_2 = (fu_2)/(u_2 - f) = ((1)(-24))/(-24-1) = (24)/(25) m ` `rArr` Speed of imageaftr1 s `= (v_1 - v_2)/(t)` ` = (29/30 - 24/25)/(1)` `= (725 - 720)/(750)` `= (5)/(750) = (1)/(150) ms^(-1)` (C)For `u_1 = -19 m` `v_1 = (fu_1)/(u_1 - f) = ((1)(-19))/(-19-1) = (19)/(20) m ` distance of person from car after 1 s, `u_2 = -19 + u=-14m` Image distance after 1 s, `v_2 = (fu_2)/(u_2 - f) = ((1)(-14))/(-14-1) = 14/15 m ` Speed of image after 1 s, ` = In (v_1 - v_2)/(t) , t = 1s ` ` = v_1 - v_2` `= (19)/(20) - (14)/(15) = (285-280)/(300) = (5)/(300)` `= (1)/(60)ms^(-1)` For `u_1 = -9M` `v_1 = (fu_1)/(u_1-f) = ((1)(-9))/(-9-1) = (9)/(10) m ` distance of person from car after 1 s, `u_2 = -9 +_ 5 = -4m` Speed of image after1 s, `v_2 = (fu_2)/(u_2 - f) = ((1)(-4))/(-4-1) = 4/5 m ` speed of image after 1 s, `= In (v_1-v_2)/(t) , t= 1 s ` `v_1 -v_2` `= 9/10 - 4/5 = (9-8)/(10) = 1/10 ms^(-1)` Here, person is moving with constant speed but it seems that its speed increasing while he comes nearer This can be experienced by person in car at rest or in moving bus This can be experienced in any moving vehicle . |
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