Suppose while sitting in a parked car,you notice a jogger approaching towards you in the rear view mirror of R=2 m.if the jogger running in a speed of 5 m/s,how fast is the image of the jogger moving,when the jogger is (a) 39 m (b) 29 m (c) 19 m and (d) 9 m away? answer are 1/280,1/150,1/60,1/10

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tenth ka level ka base par ans. dana
did u learn differentiation or not..? if we do not use differentiation, then you find, v at each point and v after one second of time.. the difference will give an approximate or crude value.
i had only learn the fomulaes 1/f=1/u+1/v
r2f
m=hi/h0=-v/u

Answers

2014-10-23T07:49:54+05:30

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If u do not know differentiation, then : we can find approx values as:

\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{u-1}{u},\ \ as\ f=1m\\ \\v=\frac{u}{u-1}\\ \\

a) u=-39m, v=\frac{-39}{-39-1}=\frac{39}{40}m\\after\ 1sec, u=-34m,v=\frac{-34}{-34-1}=\frac{34}{35}\\change\ in\ v=\frac{34}{35}-\frac{39}{40}=\frac{272-273}{280}=\frac{-1}{280}m\\

b) u=-29m, v=\frac{-29}{-29-1}=\frac{29}{30}m\\after\ 1sec, u=-24m,v=\frac{-24}{-24-1}=\frac{24}{25}\\change\ in\ v=\frac{24}{25}-\frac{29}{30}=\frac{144-145}{150}=\frac{-1}{150}m\\

c) u=-19m, v=\frac{-19}{-19-1}=\frac{19}{20}m\\after\ 1sec, u=-14m,v=\frac{-14}{-14-1}=\frac{14}{15}\\change\ in\ v=\frac{14}{15}-\frac{19}{20}=\frac{-1}{60}m\\

similarly the last one.
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If u know differentiation, then you can derive the correct solution as below.

The rear view mirror is a convex mirror.1/v + 1/u = 1/f = 2/R = 1 Differentiating both sides-\frac{1}{v^2}\frac{dv}{dt}-\frac{1}{u^2}\frac{du}{dt} = 0\\ \\\frac{dv}{dt}=-\frac{v^2}{u^2}\frac{du}{dt}\\ \\\frac{1}{v}= 1-\frac{1}{u}=\frac{u-1}{u}\\ \\v=\frac{u}{u-1}=1+\frac{1}{u-1}\\ \\\frac{dv}{dt}=\frac{1}{(u-1)^2}\frac{du}{dt}\\

du / dt = - 5 m/sec

a)  u = -39 m

\frac{dv}{dt} = -\frac{1}{40^2}*5 m/s=-0.003125 m/s

b) u = -29 m
\frac{dv}{dt} = -\frac{1}{30^2}*5 m/s=-0.0055 m/s

c) u = -19 m
\frac{dv}{dt} = -\frac{1}{20^2}*5 m/s=-0.0125m/s

d) u = -9 m
\frac{dv}{dt} = -\frac{1}{10^2}*5 m/s=-0.05m/sec 

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