The driver sitting inside a parked car is watching vehicles appro… - Vidyadip

Question

The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature $R =2 m$. Another car approaches him from behind with a uniform speed of $90 km / hr$. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the side view mirror is ' a '. The value of 100a is _________ $m / s ^2$.

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Answer

(8)
$
\begin{array}{l}Sol.\
v=\frac{u f}{u-f}=\frac{-24 \cdot 1}{-24-1}=\frac{24}{25} \\
m=\frac{-v}{u}=-\frac{24}{25(-24)}=\frac{1}{25} \\
\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\
v_1=-m^2 \cdot v_0=\frac{-1}{(25)^2} \cdot 25=\frac{-1}{25}
\\Diff.\end{array}
$
$
\begin{array}{l}
\frac{-1}{v^2}\left(\frac{d v}{d t}\right)+\frac{1}{u^2}\left(\frac{d u}{d t}\right)=0\left[\frac{d v}{d t}=v_1 ; \frac{d u}{d t}=v_0\right] \\
\frac{+2}{v^3} \cdot\left(v_1\right)^2-\frac{1}{v^2} \cdot a_1-\frac{2}{u^3} \cdot\left(v_0\right)^2+\frac{1}{u^2} \cdot a_0=0 \\
\frac{a_1}{v^2}=\frac{2}{v^3} \cdot v_1^2-\frac{2}{u^3} \cdot v_0^2 \\
a_1=\frac{2}{v} \cdot v_1^2-\frac{2 v^2}{u^3} \cdot v_0^2 \\
=\frac{2 \cdot 25}{24} \cdot \frac{1}{25} \cdot \frac{1}{25}-\frac{2}{(24)^3} \cdot \frac{24}{25} \cdot \frac{24}{25} \cdot 25 \cdot 25 \\
a_1=\frac{2}{24.25}-\frac{2}{24} \\
a_1=\frac{2}{24} \cdot \frac{-24}{25}=\frac{-2}{25} \\
100 a_1=\frac{2}{25} \times 100=8
\end{array}
$

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