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 \(5\text{ ms}^{-1}\) , how fast the image of the jogger appears to move when the jogger is \(39\text{ m }\) away.
Answer
\(\frac{1}{280}m/s\)
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Solution
Given,
Radius of curvature \(R=2\text{ m }\)
Initial distance of jogger, \(u_1=-39\text{ m }\)
After\(1s\), displacement of jogger
\(u_2=-39+5\) (As jogger is running at a speed of 5m/s)
\(=-34\text{ m }\)
Focal length,\(f=\frac R 2\)
\(=\frac 2 2\)
\(=1m\)
By mirror formulae,
\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
\(\frac 1 v=\frac 1 f-\frac 1 u\)
\(\Rightarrow v=\frac{\mathit{fu}}{u-f}\)
For \(u_1=-39\text{ m}\),
Position of image,
\(v_1=\frac{fu_1}{u_1-f}\)
\(=\frac{1\times (-39)}{-39-1}\)
\(=\dfrac{39}{40}\;\text{m}\)
For \(u_2=-34\text{ m }\),
Position of image,
\(v_2=\frac{fu_2}{u_2-f}\)
\(=\frac{1\times \left. \left(−34\right)\right. }{−34−1}\)
\(=\frac{34}{35}\)
Speed of image\(=\frac{distance\;covered}{time\;taken}\)
\(V_i=\frac{v_1-v_2} 1\)
\(=\frac{39}{40}-\frac{34}{35}\)
\(=\frac 1{280}\text{ms}^{-1}\)
Radius of curvature \(R=2\text{ m }\)
Initial distance of jogger, \(u_1=-39\text{ m }\)
After\(1s\), displacement of jogger
\(u_2=-39+5\) (As jogger is running at a speed of 5m/s)
\(=-34\text{ m }\)
Focal length,\(f=\frac R 2\)
\(=\frac 2 2\)
\(=1m\)
By mirror formulae,
\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
\(\frac 1 v=\frac 1 f-\frac 1 u\)
\(\Rightarrow v=\frac{\mathit{fu}}{u-f}\)
For \(u_1=-39\text{ m}\),
Position of image,
\(v_1=\frac{fu_1}{u_1-f}\)
\(=\frac{1\times (-39)}{-39-1}\)
\(=\dfrac{39}{40}\;\text{m}\)
For \(u_2=-34\text{ m }\),
Position of image,
\(v_2=\frac{fu_2}{u_2-f}\)
\(=\frac{1\times \left. \left(−34\right)\right. }{−34−1}\)
\(=\frac{34}{35}\)
Speed of image\(=\frac{distance\;covered}{time\;taken}\)
\(V_i=\frac{v_1-v_2} 1\)
\(=\frac{39}{40}-\frac{34}{35}\)
\(=\frac 1{280}\text{ms}^{-1}\)
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